Solucionario - Zill
January 19, 2017 | Author: MitchVillalobos | Category: N/A
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CÁLCULO INTEGRALCHAPTER 5.
298 Given f (x) =
√
INTEGRALS
x4 + 1 = (x4 + 1)1/2 , we have f (x) = 2x3 (x4 + 1)−1/2 . Thus,
f (4x) dx =
MATEMÁTICAS 2
1 1 32x3 f (4x) + C = {2(4x)3 [(4x)4 + 1]−1/2 } + C = √ + C. 4 4 256x4 + 1
16384x6 96x2 , − 256x4 + 1 (256x4 + 1)3 6x2 4x6 , we have f (4x) = − which should be the same as f (4x). Since f (x) = √ x4 + 1 (x4 + 1)3 6(4x)2 4(4x)6 96x2 16384x6 − =√ . − 256x4 + 1 (4x)4 + 1 [(4x)4 + 1]3 (256x4 + 1)3 76. First evaluating sec2 3x dx, we get To check this, take the derivative of the above function, yielding √
MANUAL DE SOLUCIONES
1 (sec2 3x)(3 dx) u = 3x, du = 3 dx 3 1 1 1 sec2 u du = tan u + C = tan 3x + C = 3 3 3 1 tan 3x + C dx, we get Next, evaluating sec2 3x dx dx = 3 1 1 tan 3x + C dx = (Cx + C1 ) + tan 3x dx 3 3 1 = (Cx + C1 ) + (tan 3x)(3 dx) u = 3x, du = 3 dx 9 1 1 tan u du = (Cx + C1 ) − ln | cos u| + C2 = (Cx + C1 ) + 9 9 1 = Cx − ln | cos 3x| + C3 . 9
sec2 3x dx =
UNIDAD 1
TEOREMA FUNDAMENTAL DEL CÁLCULO PROBLEMAS 1.1
5.3
The Area Problem
1. 3 + 6 + 9 + 12 + 15 2. −1 + 1 + 3 + 5 + 7 3. 2 + 2 + 8/3 + 4 4.
9 27 81 3 + + + 10 100 1000 10, 000
1 1 1 1 1 1 1 1 1 1 + − + − + − + 5. − + − 7 9 11 13 15 17 19 21 23 25 6. 1 −
1 1 1 1 1 1 1 1 1 + − + − + − + − 4 9 16 25 36 49 64 81 100
5.3. THE AREA PROBLEM
299
7. 0 + 3 + 8 + 15 8. 1 + 4 + 9 + 16 + 25 9. −1 + 1 − 1 + 1 − 1 10. 1 + 0 − 11.
7
1 1 +0+ 3 5
(2k + 1)
k=1
12.
6
2k
k=1
13.
13
(3k − 2)
k=1
14.
10
(4k − 2)
k=1
15.
5 (−1)k+1
k
k=1
16.
5 (−1)k k
k+1
k=1
17.
8
6
k=1
18.
9 √
k
k=1
19.
4 (−1)k+1 k=1
20.
20 k=1
22.
cos
kπ x p
5 (−1)k+1 f (k) (1) k=1
21.
k2
50 k=0
2k − 1 2k = 2
20
(x − 1)k
k=2
k=1
(−3k) = −3
50 k=1
20 · 21 2
k = −3
= 420 50 · 51 2
= −3825
CHAPTER 5. INTEGRALS
300
23.
10
(k + 1) =
k=1
24.
10 k=1
1000
(2k − 1) = 2
k=1
25.
6
(k 2 + 3) =
5
6
10
k−
k2 +
10
1=2
6
3=
k=1
5
k2 −
k=1
5
(p3 + 4) = 0 + 4 +
10
p3 +
i=1
10 i=1
29. Using Δx =
10
5 · 6 · 11 6
4=4+
p=1
i3 − 5
10
i+
i=1
6 6−0 = and f n n
− 1000 · 1 = 1, 000, 000
6 · 7 · 13 + 6 · 3 = 109 6
k=6
p=1
(2i3 − 5i + 3) = 2
1000 · 1001 2
k=1
p=0
28.
1000 k=1
k=1
k=1
27.
1000
(6k 2 − k) = 6
10 · 11 + 10 · 1 = 65 2
1=
k=1
k=1
k=1
26.
10
k+
10
−
102 · 112 + 10 · 4 = 3069 4
3=2
i=1
a+k
b−a n
5·6 = 315 2
102 · 112 4
=f
6k n
=
−5
10 · 11 2
+ 10 · 3 = 5805
6k we have n
n n 6k 6 36 36 n(n + 1) 1 A = lim = lim = lim 18 1 + k = lim 2 · = 18. n→∞ n→∞ n n→∞ n n n→∞ n2 2 n k=1
30. Using Δx =
k=1
2 3−1 = and f n n
a+k
b−a n
=f
1+
2k n
=2+
4k we have n
n n n 4 2 8k 4 8 = lim + A = lim 1+ 2 k = lim n→∞ n→∞ n n→∞ n n2 n n k=1 k=1 k=1 k=1
8 n(n + 1) 4 1 ·n+ 2 · = lim = lim 4 + 4 1 + = 8. n→∞ n n→∞ n 2 n n
4 31. Using Δx = and f n
4k 2+ n
b−a a+k n
=3+
8k we have n
n n n 12 32k 4 32k 12 = lim + 2 = lim A = lim 1+ 2 k n→∞ n→∞ n n→∞ n n n n k=1 k=1 k=1 k=1
12 32k n(n + 1) 1 ·n+ 2 · = lim = lim 12 + 16 1 + = 28. n→∞ n n→∞ n 2 n n
8k 3+ n
5.3. THE AREA PROBLEM 2 32. Using Δx = and f n
301
b−a a+k n
=
6k we have n
n n 6k 2 12 12 n(n + 1) 1 = lim = lim 6 1 + A = lim k = lim 2 · = 6. n→∞ n→∞ n n→∞ n n n→∞ n2 2 n k=1
k=1
2 33. Using Δx = and f n
b−a a+k n
n 4k 2
=
4k 2 we have n2
n 8 2 8 n(n + 1)(2n + 1) k = lim · A = lim = lim 3 n→∞ n→∞ n n→∞ n3 n2 6 k=1 k=1 1 4 8 3 2+ + 2 = . = lim n→∞ 3 n n 3 34. Using Δx =
3 and f n
a+k
b−a n
=4−
12k 9k 2 + 2 we have n n
n 3 12 36 27k 2 = lim − 2+ 3 n→∞ n n→∞ n n n k=1 k=1 n n n 12 36 27 2 1− 2 k+ 3 k = lim n→∞ n n n k=1 k=1 k=1
12 36 n(n + 1) 27 n(n + 1)(2n + 1) ·n− 2 · + 3· = lim n→∞ n n 2 n 6
1 1 9 3 = lim 12 − 18 1 + = 12 − 18 + 9 = 3. + 2+ + 2 n→∞ n 2 n n
A = lim
n
2 · n
4−
12k 9k 2 + 2 n n
4k 4k 2 − 2 we have n n n n n 8 2 4k 4k 2 2 8 A = lim − 2 = lim k− 3 k n→∞ n n n n→∞ n2 n k=1 k=1 k=1
1 4 1 8 4 3 = lim 4 1 + =4− = . − 2+ + 2 n→∞ n 3 n n 3 3
35. Using Δx =
2 and f n
36. Using Δx =
2 and f n
a+k
b−a n
a+k
b−a n
=
=f
−3 +
2k n
= 21 −
24k 8k 2 + 2 we have n n
n n n 48 16 2 42 24k 8k 2 2 A = lim − 2 = lim 1− 2 k+ 3 k 21 − n→∞ n n n n→∞ n n n k=1 k=1 k=1 k=1
1 70 1 16 8 3 = . = lim 42 − 24 1 + = 42 − 24 + + 2+ + 2 n→∞ n 3 n n 3 3 n
CHAPTER 5. INTEGRALS
302
b−a k2 4k a+k + 2 we have =3+ n n n n n n n k2 1 3 4 1 2 4k + 2 = lim 1+ 2 k+ 3 k 3+ A = lim n→∞ n n n n→∞ n n n k=1 k=1 k=1 k=1
1 16 1 1 1 3 . = lim 3 + 2 1 + =3+2+ = + 2+ + 2 n→∞ n 6 n n 3 3
1 37. Using Δx = and f n
b−a 4k 4k 2 a+k + 1 we have = 2 − n n n n n n n 2 4k 2 4k 8 2 8 2 +1 = lim − k − 2 k+ 1 A = lim n→∞ n2 n n n→∞ n3 n n k=1 k=1 k=1 k=1
4 1 2 1 3 8 = lim 2+ + 2 −4 1+ +2 = −4+2= . n→∞ 3 n n n 3 3
2 38. Using Δx = and f n
b−a k3 = 3 we have n n n n k3 1 1 1 3 1 1 2 · k = lim A = lim = lim 4 1+ + 2 = . 3 n→∞ n→∞ n→∞ n n n 4 n n 4
39. Using Δx =
1 and f n
a+k
k=1
k=1
2 b−a 12k 2 8k 3 and f a + k = 3 − 2 + 4 we have n n n n n n n n 2 8k 3 12k 2 16 3 24 2 8 = lim − 2 +4 k − 3 k + 1 A = lim n→∞ n3 n n n→∞ n4 n n k=1 k=1 k=1 k=1
1 1 2 3 = lim 4 1 + + 2 − 4 2 + + 2 + 8 = 4 − 8 + 8 = 4. n→∞ n n n n
40. Using Δx =
1) and A2 is the area under 41. Let A = A1 + A2 where A1 is the area under f (x) = 2 on [0, b−a 1 f (x) = x + 1 on [1, 4]. For A1 , we have Δx = , f a + k = 2, and n n n n 2 2n 1 = 2. 1 = lim 2· = lim A1 = lim n→∞ n→∞ n→∞ n n n k=1 k=1 b−a 3 3k , and For A2 , we have Δx = , f a + k =2+ n n n n n n 9 6 3k 3 A2 = lim = lim 1+ 2 k 2+ n→∞ n n n→∞ n n k=1 k=1 k=1
6n 9 21 1 9 + . = lim 1+ =6+ = n→∞ n 2 n 2 2
5.3. THE AREA PROBLEM Then A = 2 +
303
25 21 = . 2 2
42. Let A = A1 + A2 where A1 is the area under f (x) =−x + 1 on [0, 1) and A2 is the area under b−a 1 k f (x) = x + 2 on [1, 3]. For A1 , we have Δx = , f a + k = 1 − , and n n n
A1 = lim
n
n→∞
k=1
k 1− n
For A2 , we have Δx =
A2 = lim
n→∞
Then A =
n k=1
1 = lim n n→∞
2 ,f n
2k 3− n
a+k
n n 1 1 1− 2 k n n k=1
b−a n
2 = lim n n→∞
=3+
= lim
n→∞
k=1
1 1− 2
1 1+ n
=
1 . 2
2k , and n
n n 4 6 1+ 2 k n n k=1
k=1
= lim
n→∞
1 6+2 1+ = 10. n
17 1 +8= . 2 2 3
77 1 2 1 1 1 2 1 + · + · + · = 2 3 2 2 2 5 2 60 1 2 1 1 1 25 1 AL = 2 · + 1 · + · + · = 2 2 3 2 2 2 12
43. AR = 1 ·
2 1 1
√
2 π · +1· 44. AR = 2 4 √ 2 π · AL = 0 · + 4 2
√ 2 π π + · +0· 4 2 4 √ π π 2 +1· + · 4 4 2
√ π (1 + 2)π = 4 4√ π (1 + 2)π = 4 4
2
3
1
2
2 -1
CHAPTER 5. INTEGRALS
304
3 2 − (−1) 3 = and x∗k = −1 + (k − 1) we obtain n n n
2 n n 3 3 ∗ A = lim f (xk )Δx = lim 4 − −1 + (k − 1) n→∞ n→∞ n n k=1 k=1
n (k − 1)2 3 k−1 −9 = lim 3+6 n→∞ n n n2 k=1
n n n 3 6 9 2 3 1+ (k − 1) − 2 (k − 2k + 1) = lim n→∞ n n n k=1 k=1 k=1
n n n n n n 3 6 6 9 2 18 9 3 = lim 1+ k− 1− 2 k + 2 k− 2 1 n→∞ n n n n n n k=1 k=1 k=1 k=1 k=1 k=1
9 18 n(n + 1) 18 27 n(n + 1)(2n + 1) 54 n(n + 1) 27 n+ 2 − 2n − 3 + 3 − 3n = lim n→∞ n n 2 n n 6 n 2 n
1 18 9 1 1 27 1 27 = lim 9 + 9 1 + − − 1+ 2+ + 1+ − 2 n→∞ n n 2 n n n n n
45. Using Δx =
= 9 + 9 − 0 − 9 + 0 − 0 = 9. 3 2 − (−1) 2k − 1 3 = and x∗k = −1 + we obtain n n 2 n
2 n n 2k − 1 3 3 ∗ A = lim 4 − −1 + f (xk )Δx = lim n→∞ n→∞ 2 n n k=1 k=1
n 3 2k − 1 9 (2k − 1)2 − = lim 3+3 n→∞ n n 4 n2 k=1
n n n 3 3 9 2 3 1+ (2k − 1) − 2 (4k − 4k + 1) = lim n→∞ n n 4n k=1 k=1 k=1
n n n n n n 3 6 3 9 2 9 9 3 = lim 1+ k− 1− 2 k + 2 k− 2 1 n→∞ n n n n n 4n k=1 k=1 k=1 k=1 k=1 k=1
9 18 n(n + 1) 9 9 27 n(n + 1)(2n + 1) 27 n(n + 1) n+ 2 − 2n − 3 + 3 − 3n = lim n→∞ n n 2 n n 6 n 2 4n
9 1 9 1 1 27 1 9 = lim 9 + 9 1 + − − 1+ 2+ + 1+ − 2 n→∞ n n 2 n n 2n n 4n
46. Using Δx =
= 9 + 9 − 0 − 9 + 0 − 0 = 9.
5.3. THE AREA PROBLEM
305
b−a 47. Identify b − a = 2. Taking a = 0, we have f a + k = n 2 √ 2k 2k . Then A is the area under f (x) = 4 − x2 = 4− f n n from x = 0 to x = 2.
2
-2
2 -2
b−a 48. Identify b − a = π. Taking a = 0, we have f a + k = n kπ kπ . Then A is the area under f (x) = sin x from f = sin n n x = 0 to x = π.
2
-
-2
8
1 1 1 1 + 2 + ··· + 8 = 10 10 10 10k
49. 0.11111111 =
k=1
37 37 37 + 50. 0.3737373737 = + ··· + = 37 100 1002 1005
51.
60
k2 =
k=21
52.
6
60
k2 −
k=1
k=0 n
xk −
n
n
8
(k + 5);
k=1
x=
1 1 1 + + ··· + 100 1002 1005
= 37
5 k=1
1 100k
60 · 61 · 121 20 · 21 · 41 − = 70, 940 6 6
k=1
k=1
54. (a)
k2 =
k=1 7
(k + 6);
53. 0 =
20
(k + 4)
k=2 n
1 xk n n
xk − nx;
x=
k=1
k=1
[f (k) − f (k − 1)] = [f (1) − f (0)] + [f (2) − f (1)] + [f (3) − f (2)] + · · ·
k=1
+ [f (n − 1) − f (n − 2)] + [f (n) − f (n − 1)] = f (n) − f (0) (b) f (k) =
√
k;
400 √ √ √ √ ( k − k − 1) = 400 − 0 = 20 k=1
55. (a) Identifying f (k) = (k + 1)2 in part (a) of Problem 54, we have
k=1
(n + 1)2 − 12 = n2 + 2n. (b)
n k=1
[(k + 1)2 − k 2 ] =
n k=1
(2k + 1) =
n
n k=1
2k +
n k=1
1=2
n k=1
k+n
[(k + 1)2 − k 2 ] =
CHAPTER 5. INTEGRALS
306
(c) Comparing the results of (a) and (b), we find that equating them leads to summation formula (ii):
2
n
n
k + n = n2 + 2n;
k=1
k=1
k=
n2 + n n(n + 1) n2 + 2n − n = = 2 2 2
Using f (k) = (k + 1)3 similarly to (a), we obtain n
[(k + 1)3 − k 3 ] = (n + 1)3 − 13 = n3 + 3n2 + 3n.
k=1
Analogously for (b), we also have n
[(k + 1)3 − k 3 ] =
k=1
n
(3k 2 + 3k + 1) = 3
k=1
n
k2 + 3
k=1
n
k + n.
k=1
Combining these, we obtain
3
n
k2 + 3
k=1
3
n
k2 +
k=1
n
k + n = n3 + 3n2 + 3n
k=1
3n(n + 1) + n = n3 + 3n2 + 3n 2 3
n k=1 n
k 2 = n3 + 3n2 + 2n − k2 =
k=1
=
3n2 + 3n 2
2n3 + 3n2 + n 2n3 + 6n2 + 4n − 3n2 − 3n = 6 6 n(n + 1)(2n + 1) n(2n2 + 3n + 1) = . 6 6
56. The pattern illustrated in Figure 1.1.9 5.3.9 indicates that the summation of cubes is the square n 2 n 3 of the summation of the numbers being cubed. That is: k = k . Expanding the summation, we get
n k=1
k3 =
n k=1
2 k
=
n(n + 1) 2
k=1
2
2
=
k=1 2
n (n + 1) . 4
5.3. THE AREA PROBLEM
307
h2 − h1 x + h1 . 57. The equation of the line through (0, h1 ) and (b, h2 ) is f (x) = b b−a kb b k(h2 − h1 ) + h1 we find Using Δx = and f a + k =f = n n n n
n n n k(h1 − h2 ) b bh1 b + h1 = lim A = lim (h2 − h1 ) k+ 1 n→∞ n n n→∞ n2 n k=1 k=1 k=1
b(h2 − h1 ) 1 bh1 n b(h2 − h1 ) + bh1 = lim 1+ + = n→∞ 2 n n 2 h1 + h2 bh2 − bh2 + 2bh1 = = b. 2 2
y
(b, h2) h2
h1 b
58. Since the total number of cans is 136 and there is one additional can per row, we have n n(n + 1) k= = 136. Solving for n will yield the number of cans in the bottom row, so 2 k=1 we have n2 + n − 272 = 0 and (n − 16)(n + 17) = 0, yielding n = 16 or n = −17. Thus, there are 16 cans in the bottom row.
4 59. Using Δx = and f n
A = lim
n→∞
n 128k k=1
n
−
b−a a+k n
=
512k 3 256k 4 128k 384k 2 − + − we have n n2 n3 n4
384k 2 512k 3 256k 4 + − 2 3 n n n4
4 n
n n n n 512 1536 2 2048 3 1024 4 = lim k− 3 k + 4 k − 5 k n→∞ n2 n n n k=1 k=1 k=1 k=1
1 1 1 1 2 512 3 15 10 + 2− 4 = lim 256 1 + − 256 2 + + 2 + 512 1 + + 2 − 6+ n→∞ n n n n n 15 n n n 256 1024 = . = 256 − 512 + 512 − 5 5
2 60. We note that A2 = 1 − A 1 where A1is the2 area under y = x from 0 to b−a 1 k 1. Using Δx = and f a + k = 2 we find n n n
n n 2 k 1 1 2 1 1 1 3 + k = lim = . = lim 2 + 2 3 2 n→∞ n→∞ n n→∞ 6 n n n n 3
A1 = lim
k=1
Thus, A2 = 1 −
k=1
2 1 = . 3 3
1
A1 A2 1
CHAPTER 5. INTEGRALS
308
61. We note that A2 = 16 − A1 where A1is the area under y = x3 from 0 to b−a 2 8k 3 2. Using Δx = and f a + k = 3 we find n n n A1 = lim
n→∞
n k=1
3
8k 2 n3 n
= lim
n→∞
Thus, A2 = 16 − 4 = 12. x0 and f 62. (a) Using Δx = n
16 n4
n
k 3 = lim 4 1 + n→∞
k=1
b−a a+k n
=f
kx0 n
1 2 + n n2
=a
2
A1
= 4.
A2 8
kx0 k 2 x20 + c we have +b 2 n n
n n n n x0 kx0 x20 x0 x30 2 k 2 x20 +c = lim a 3 k +b 2 k+c 1 A = lim a 2 +b n→∞ n→∞ n n n n n n k=1 k=1 k=1 k=1 3
1 x2 x x2 x3 3 1 = lim a 0 2 + + 2 + b 0 1 + + cx0 = a 0 + b 0 + cx0 . n→∞ 6 n n 2 n 3 2 (b) Let A1 be the area under the graph on [0, 2] and A2 the area under the graph on [0, 5]. Then the area under the graph on [2, 5] is 3 3 52 2 22 5 A = A2 − A1 = 6 + 2 + 1.5 − 6 + 2 + 1.2 3 2 3 2 = (250 + 25 + 5) − (16 + 4 + 2) = 258. 63. By (8) of this section, n n 1 1 1 = lim f 0 + [k − 1] e(k−1)/n · n→∞ n n n→∞ n k=1 k=1 1 = lim 1 + e1/n + e2/n + · · · + e(n−1)/n n→∞ n 1 1 + e1/n + (e1/n )2 + · · · + (e1/n )n−1 . = lim n→∞ n
A = lim
Using a = 1, r = e1/n , we obtain
1 − (e1/n )n 1 1 A = lim ·1· = (1 − e) lim 1/n n→∞ n n→∞ 1−e n(1 − e1/n ) 1 − e1/n n→∞ 1/n
Now, lim n(1 − e1/n ) = lim n→∞
h
−e1/n (−1/n2 ) = − lim e1/n = −1, n→∞ n→∞ −1/n2
= lim so A = (1 − e)
1 −1
(form ∞ · 0)
= e − 1.
5.4. THE DEFINITE INTEGRAL 64. 1 + 3 + 5 + · · · + 2n − 1 =
n
309 (2k − 1) = 2
k=1
n k=1
k−
n
1 = n(n + 1) − n = n2
k=1
The total distance moved is thus proportional to 1 + 3 + 5 + · · · + 2n − 1 = n2 .
PROBLEMAS 1.2
5.4
The Definite Integral
1. From Δx1 = 1, Δx2 = 2/3, Δx3 = 2/3, and Δx4 = 2/3 we see that the norm of the partition is P = 1. Using f (x∗1 ) = 5/2, f (x∗2 ) = 5, f (x∗3 ) = 7, and f (x∗4 ) = 9 we compute the Riemann sum 4 2 2 2 5 33 ∗ . f (xk )Δxk = (1) + 5 +7 +9 = 2 3 3 3 2 k=1
2. From Δx1 = 1, Δx2 = 1/2, Δx3 = 1, Δx4 = 5/2, and Δx5 = 2 we see that the norm of the partition is P = 5/2. Using f (x∗1 ) = −11/2, f (x∗2 ) = −9/2, f (x∗3 ) = −4, f (x∗4 ) = −2, and f (x∗5 ) = 0 we compute the Riemann sum 5 1 5 11 9 67 ∗ f (xk )Δxk = − (1) + − + (−4)(1) + (−2) + 0(2) = − . 2 2 2 2 4 k=1
3. From Δx1 = 3/4, Δx2 = 1/2, Δx3 = 1/2, and Δx4 = 1/4 we see that the norm of the partition is P = 3/4. Using f (x∗1 ) = 9/16, f (x∗2 ) = 0, f (x∗3 ) = 1/4, and f (x∗4 ) = 49/64 we compute the Riemann sum 4 1 9 3 1 1 49 1 189 ∗ . f (xk )Δxk = +0 + + = 15 4 2 4 2 64 4 256 k=1
4. From Δx1 = 1/2, Δx2 = 1, and Δx3 = 1/2 we see that the norm of the partition is P = 1. Using f (x∗1 ) = 41/16, f (x∗2 ) = 65/16, and f (x∗3 ) = 10 we compute the Riemann sum 3 1 41 1 65 331 ∗ . f (xk )Δxk = + (1) + 10 = 16 2 16 2 32 k=1
5. From Δx1 = π, Δx2 = π/2, and Δx3 = π/2 we see √ that the norm of the partition is P = π. Using f (x∗1 ) = 1, f (x∗2 ) = −1/2, and f (x∗3 ) = − 2/2 we compute the Riemann sum √ √ 3 π π 2 1 (3 − 2)π ∗ . f (xk )Δxk = 1(π) + − + − = 2 2 2 2 4 k=1
6. From Δx1 = π/4, Δx2 = π/4, Δx3 = π/3, and Δx√ 4 = π/6 we see √that the norm of the partition is P = π/3. Using f (x∗1 ) = 1/2, f (x∗2 ) = 3/2, f (x∗3 ) = 2/2, and f (x∗4 ) = 1/2 we compute the Riemann sum √ √ √ √ 4 3 π 2 π 1 π (5 + 3 3 + 4 2)π 1π ∗ + + + = . f (xk )Δxk = 2 24 2 4 2 3 2 6 24 k=1
CHAPTER 5. INTEGRALS
310
7. We have Δxk = 1 and x∗k = k for k = 1, 2, 3, 4, 5. Using f (x∗1 ) = −1, f (x∗2 ) = 0, f (x∗3 ) = 1, f (x∗4 ) = 2, and f (x∗5 ) = 3 we compute the Riemann sum 5
f (x∗k )Δxn = −1(1) + 0(1) + 1(1) + 2(1) + 3(1) = 5.
k=1
k−1 for k = 1, 2, 3. Using f (x∗1 ) = 1, f (x∗2 ) = 7/9, and 3 f (x∗3 ) = 7/9 we compute the Riemann sum 3 1 7 1 7 1 23 ∗ . f (xk )Δxn = 1 + + = 3 9 3 9 3 27
8. We have Δxk = 1/3 and x∗k =
k=1
4
9.
9 + x2 dx
−2 π/4
10.
tan x dx 0
11. Identify a = 0 and b = 2. Then
b−a b−a b−a 2k 2k 2 2k = f a+k and f a + k . 1+ =f =1+ n n n n n n n 2 n 2k 2 = (x + 1) dx. 1+ Taking f (x) = x + 1 we have lim n→∞ n n 0 k=1
12. Identify a = 1 and b = 4. Then 3
3 3 b−a b−a b−a 3k 3k 3k = f a+k and f a + k . 1+ =f 1+ = 1+ n n n n n n n 3 4 n 3 3k = x3 dx. 1+ Taking f (x) = x3 we have lim n→∞ n n 1 k=1 4 b−a b−a 4k = and f a + k we have 13. Using = −3 + n n n n 1 n n n 16 12 4k 4 = lim − x dx = lim 1+ 2 k −3 + n→∞ n n n→∞ n n −3 k=1 k=1 k=1
1 12n +8 1+ = lim − = −12 + 8 = −4. n→∞ n n 3 b−a 3k b−a = and f a + k we have = 14. Using n n n n
3 n n 3k 3 9 9 1 9 · x dx = lim k = lim = lim 2 1+ = . n→∞ n→∞ n→∞ n n n 2 n 2 0 k=1
k=1
5.4. THE DEFINITE INTEGRAL 15. Using
k2 k k − 1+ = + 2 we have n n n 2 n n n k k2 1 1 2 1 2 + = lim (x − x) dx = lim k+ 3 k n→∞ n n2 n n→∞ n2 n 1 k=1 k=1 k=1
1 1 5 1 1 1 1 3 = lim = + = . 1+ + 2+ + 2 n→∞ 2 n 6 n n 2 3 6
b−a 1 = and f n n
5 b−a 16. Using = and f n n
311
a+k
b−a n
b−a a+k n
=
1+
=
k n
2
5k −2 + n
2 −4=−
20k 25k 2 + 2 we have n n
n n n 100 125 2 20k 25k 2 5 + 2 = lim − 2 (x − 4) dx = lim k+ 3 k − n→∞ n n n n→∞ n n −2 k=1 k=1 k=1
1 25 100 125 1 125 3 =− . = lim − = −50 + 1+ + 2+ + 2 n→∞ 2 n 6 n n 3 3 3
2
b−a 1 17. Using = and f n n
1 0
2 0
19. Using f
b a
b−a a+k n
=
k3 − 1 we have n3
n n 1 1 3 1 = lim (x − 1) dx = lim −1 k − 1 n→∞ n3 n n→∞ n4 n k=1 k=1 k=1
1 1 3 n 2 1 = lim 1+ + 2 − = −1=− . n→∞ 4 n n n 4 4 n 3 k
3
b−a 2 18. Using = and f n n
b−a a+k n
=3−
8k 3 we have n3
n n 6 16 3 8k 3 2 = lim (3 − x ) dx = lim 1− 4 k 3− 3 n→∞ n n n→∞ n n k=1 k=1 k=1
1 2 = lim 6 − 4 1 + + 2 = 6 − 4 = 2. n→∞ n n n
3
b−a a+k n
=a+
k(b − a) we have n
n n (b − a)2 a(b − a) k(b − a) b − a = lim x dx = lim 1+ k a+ n→∞ n→∞ n n n n2 k=1 k=1 k=1
a(b − a)n (b − a)2 1 (b − a)2 = lim + 1+ = a(b − a) + n→∞ n 2 n 2 b−a b2 − a 2 b−a (2a + b − a) = (b + a) = . = 2 2 2 n
CHAPTER 5. INTEGRALS
312 20. Using f
a+k
b
b−a n
2 2ka(b − a) k 2 (b − a)2 k(b − a) + = a2 + we have = a+ n n n2 n
2
x dx = lim
n→∞
a
3
21.
x dx = −1
3 −1
6 3 5
24. −2
1 2 [3 − (−1)2 ] = 4 2 1 3 28 [3 − (−1)3 ] = 3 3
(−2) dx = −2(5 + 2) = −14
−2
25. 4
4 dx = 4(6 − 3) = 12
23.
n n n a2 (b − a) 2a(b − a)2 (b − a)3 2 = lim 1+ k+ k n→∞ n n2 n3 k=1 k=1 k=1 2
a (b − a)n 2a(b − a)2 1 1 (b − a)3 3 = lim + 1+ + 2+ + 2 n→∞ n 2 n 6 n n 3 b−a 2 (b − a) = [3a + 3a(b − a) + (b − a)2 ] = a2 (b − a) + a(b − a)2 + 3 3 b−a 2 b3 − a 3 = (b + ab + a2 ) = . 3 3
x2 dx =
22.
k=1
2ka(b − a) k 2 (b − a)2 b − a + a + n n2 n 2
5
1 1 dx = (−2 − 4) = −3 2 2
10x4 dx = 0
26. 5
27. −
−1
3
10x dx = 10
x dx = 10(4) = 40
3
−1
3
28.
3
(3x + 1) dx = −1
3
−1
−1
29. 3
30. −1
t2 dt = − 2
3 −1
(3x − 5) dx =
3
3x dx + −1
t2 dt = −
3 −1
2
1 dx = 3(4) + 1[3 − (−1)] = 16
28 3
3x dx −
3
5 dx = 3 −1
28 3
− 5[3 − (−1)] = 8
5.4. THE DEFINITE INTEGRAL
3
31. −1
(−3x2 + 4x − 5) dx = −3
313
3
x2 dx + 4
−1
3 −1
x dx −
3
5 dx −1
28 + 4(4) − 5[3 − (−1)] = −32 3 3 3 3 3 28 2 2 32. 6x(x − 1) dx = (6x − 6x) dx = 6 x dx − 6 x dx = 6 − 6(4) = 32 3 −1 −1 −1 −1 3 3 0 28 33. x2 dx + x2 dx = x2 dx = 3 −1 0 −1 1.2 1.2 1.2 3 3 3 34. 2t dt − 2t dt = 2t dt + 2t dt = 2t dt = 2 t dt = 2(4) = 8 = −3
−1
35.
3
4
4
x dx +
0
0
0
36.
−1 3
37.
t2 dt +
−1
(9 − x) dx =
2
x2 dx +
0
x3 dx +
0
0
t3 dt =
3 −1
38. −1
−1 3
0
3
[x + (9 − x)] dx =
u2 du =
2
0
5x dx −
3
1.2
4
x3 dx −
0
x2 dx +
3
4 0
−1
−1
2
(x − 4) dx = 0 +
9 dx = 9(4 − 0) = 36
x2 dx +
0
3
x2 dx =
2
3
x2 dx =
−1
28 3
x3 dx = 0
0
−1
3 −1
(x − 4) dx =
3 −1
x dx −
3
4 dx −1
= 4 − 4[3 − (−1)] = −12 5 2 5 39. f (x) dx = f (x) dx − f (x) dx = 8.5 − 6 = 2.5
2
3
40.
0
4
f (x) dx =
1
1
f (x) dx −
2
41.
0
4 3
f (x) dx = 2.4 − (−1.7) = 4.1
−1
1 = 2(3.4) + 3
2 −2
2
2f (x) dx +
−1
42. Since
2
[2f (x) + g(x)] dx =
g(x) dx = 2 −1
2
1 f (x) dx + (3) 3 −1
2
g(x) dx −1
2
1 3g(x) dx = 6.8 + (12.6) = 6.8 + 4.2 = 11 3 −1
[f (x) − 5g(x)] dx = 24, we have
2 −2
f (x) dx − −5
2
5g(x) dx = 24 −2 2 −2 2 −2
g(x) dx = 24 − g(x) dx = −
2
f (x) dx −2
−2 38 1 24 + 14 =− . f (x) dx = − 24 + 5 5 5 2
CHAPTER 5. INTEGRALS
314
b
43. (a)
a
b
c
f (x) dx = −2.5
c
(b)
f (x) dx = 3.9 d
(c)
f (x) dx = −1.2
c
(d)
f (x) dx =
a
b
c
f (x) dx +
a
f (x) dx = 3.9 − 1.2 = 2.7
c
b
f (x) dx =
d
f (x) dx +
d
(f)
c
f (x) dx = b
f (x) dx = −2.5 + 3.9 = 1.4
b
d
c
f (x) dx + a
(e)
b
a
b
f (x) dx = −2.5 + 3.9 − 1.2 = 0.2
c
b
44. (a)
f (x) dx = 6.8
a
b
c
c
(b)
f (x) dx = −7.3
d
(c)
f (x) dx = 9.2
c
(d) a
(e)
b
b
f (x) dx = a
c
d
f (x) dx +
d
f (x) dx = 6.8 − 7.3 = −0.5
b
f (x) dx = b
c
f (x) dx + a
d
(f)
b
f (x) dx =
f (x) dx = −7.3 + 9.2 = 1.9
c
c
f (x) dx + a
b
6
4
4
2
2 2
-2
4
f (x) dx = 6.8 − 7.3 + 9.2 = 8.7
c
47.
6
-2
d
f (x) dx +
46.
45.
-4
d
f (x) dx +
-4
-2
48. 1 2
2
4
-2
2
3
-1
-2
2
49. From the figure, we see that the area under the graph is a triangle with a base and height of 6. Thus, the area from geometry is
6 4
6(6) bh = = 18. A= 2 2
2 -4
-2 2
2
4
5.4. THE DEFINITE INTEGRAL
315
50. From the figure, we see that the area under the graph consists of two triangles; one has a base and height of 1 while the other has a base and height of 2. Thus, the area from geometry is
4
2
b2 h2 1(1) 2(2) 5 b 1 h1 + = + = . A= 2 2 2 2 2
2
51. From the figure, we see that the area under the graph consists of onefourth of a circle of radius 3. Thus, the area from geometry is 2
A=
4
4
2
π(3) 9π πr = = . 4 4 4
2
-2
2
52. From the figure, we see that the area under the graph consists of a semicircle of radius 3 above a rectangle of width 6 and height 2. Thus, the area from geometry is A=
53.
4 2
π(3)2 9π πr2 + wh = + 6(2) = + 12. 2 2 2
54.
-2
55.
56.
2
6
2
2 2
4 2
-2 2
-2
4
2
-2
6 -2
-4
2
4
2
-2 -4
-2
57. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 1 and a height of 2 subtracted from the area of a triangle with a base of 4 and a height of 8. Thus, the net signed area from geometry is
8 6 4 2
b2 h2 4(8) 1(2) b 1 h1 − = − = 15. A= 2 2 2 2 58. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 4 and a height of 2 subtracted from the area of a triangle with a base of 4 and a height of 2. Thus, the net signed area from geometry is
-2
2
2
2
b2 h 2 4(2) 4(2) b 1 h1 − = − = 0. A= 2 2 2 2
4
-2
-2
4
6
8
CHAPTER 5. INTEGRALS
316
1
59. −1
(x −
1 − x2 ) dx can be rewritten as
1 −1
x dx −
1
−1
1 − x2 dx, so the net signed area
of the graph below left is the same as the difference between the net signed areas of the graphs below right. This difference, in turn, is the area of a semicircle of radius 1 subtracted from the net signed area of two triangles with bases and heights of 1. From geometry, this is
1(1) 1(1) b 1 h1 b2 h2 π πr2 π(1)2 − = − =− . A= − − 2 2 2 2 2 2 2 2
-2
2
2
-2
=
-2
2
2
−
-2
-2
2
-2
60. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 1 and a height of 1 subtracted from the area of a triangle with a base of 2 and a height of 1. Thus, the net signed area from geometry is
2
-2
2
b2 h 2 2(1) 1(1) 1 b1 h1 − = − = . A= 2 2 2 2 2
-2
61. From the figure, we see that the net signed area under the graph is the negative of the area of a triangle with a base of 2 and a height of 2. Thus, the net signed area from geometry is A=−
2(2) bh =− = −2. 2 2
3
-3
3 -3
62. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 1 and a height of 1 subtracted from the area of a triangle with a base of 3 and a height of 3. Thus, the net signed area from geometry is
3
-3
b2 h2 3(3) 1(1) b 1 h1 − = − = 4. A= 2 2 2 2 63. From the figure, we see that the net signed area under the graph is the area of a triangle with a base of 4 and a height of 4 subtracted from the sum of the areas of a triangle with a base of 3 and a height of 3, and a rectangle of width 2 and height 3. Thus, the net signed area from geometry is
3 -3
3
-3
3 -3
b 2 h2 3(3) 4(4) 5 b 1 h1 )− = [2(3) + ]− = = 2.5. A = (wh + 2 2 2 2 2
5.4. THE DEFINITE INTEGRAL
317
64. From the figure, we see that the net signed area under the graph is the sum of the areas of a triangle with a base of 3 and a height of 3, and a rectangle of width 7 and height 3. Thus, the net signed area from geometry is
3
3
3(3) 51 bh = 7(3) + = = 25.5. A = wh + 2 2 2
6
9
-3
65. For −1 ≤ x ≤ 0, ex ≤ 1 and e−x ≥ 1. Then e−x ≥ ex on [−1, 0] and by Theorem 5.4.7(i) we 0 0 have ex dx ≤ e−x dx. −1
−1
66. For 0 ≤ x ≤ π/4, tan x ≤ 1. Then sin x/ cos x ≤ 1 and sin x ≤ cos x. Thus, by Theorem 5.4.7(i) we have
π/4 0
π/4
and 0
cos x dx ≥
π/4
π/4
cos x dx −
sin x dx, 0
0
π/4 0
sin x dx ≥ 0,
(cos x − sin x) dx ≥ 0.
3 2 3 x (x + 1)−1/2 . For 0 ≤ x ≤ 1, f (x) ≥ 0 and 2 √ f (0) ≤ f (x) ≤ f (1). Since f (0) = 1 and f (1) = 2 < 1.42, we identify m = 1 and M = 1.42. Then by Theorem 1.2.7 5.4.7(ii) (ii)
67. Letting f (x) = (x3 + 1)1/2 we have f (x) =
1(1 − 0) ≤
1 0
(x3 + 1)1/2 dx ≤ 1.42(1 − 0)
and
1≤
1 0
(x3 + 1)1/2 dx ≤ 1.42.
68. Letting f (x) = x2 − 2x we have f (x) = 2x − 2 and f (x) = 2. Solving f (x) = 0 we obtain the critical number 1, and since f (x) > 0 for all x, the graph of f is concave up with the absolute minimum at x = 1. Since f (0) = f (2) = 0, we identify m = −1 and M = 0. Then by Theorem 5.4.7(ii) 1.2.7 (ii) −1(2 − 0) ≤
2
3
2 0
(x2 − 2x) dx ≤ 0(2 − 0)
2
2
and
−2≤
3
2 0
(x2 − 2x) dx ≤ 0.
69. On [0, 1], x −x = x (1−x) ≥ 0, so x ≥ x . Thus by Theorem 5.4.7(i), 70. On [0, 1], x2 −x = x(x−1) ≤ 0, so x2 ≤ x. Thus, 1 1 √ 2 4 + x dx ≤ 4 + x dx. 0
√
4 + x2 ≤
0
2
71. Since f (x) ≥ 0 on [a, b], by (12),
b a
f 2 (x) dx ≥ 0.
√
1 0
2
x dx ≥
1
x3 dx.
0
4 + x, and by Theorem 5.4.7(i), 1.2.7 (i)
CHAPTER 5. INTEGRALS
318
72. We will use the fact that any interval with nonzero length contains both rational and irra 1 n f (x) dx = lim f (x∗k )Δxk . tional numbers. Let P be a partition of [−1, 1]. Then First, choosing each
x∗k
choosing each
Since 0 = 2,
f (x) dx = lim
to be rational, we obtain
x∗k
to be irrational, we obtain
P →0
−1
n
1 P →0
−1
1
f (x) dx = lim −1
1
P →0
n
k=1
0 · Δxk = 0. Then,
k=1
1 · Δxk = 1 − (−1) = 2.
k=1
f (x) dx does not exist. −1
k2 (k − 1)2 2k − 1 − = we have 2 n n2 n2 1 n n √ k 2 2k − 1 1 2 x dx = lim (2k − k) = lim 3 n→∞ n→∞ n n2 n2 0 k=1 k=1
n n 2 n(n + 1)(2n + 1) 1 n(n + 1) 1 2 2 − = lim k − k = lim · · n→∞ n→∞ n3 n3 n3 6 n3 2 k=1 k=1
1 2 1 1 1 1 2 = lim 1+ 2+ − 1+ = −0= . n→∞ 3 n n 2n n 3 3
73. Using Δx =
π 1 π (2k − 1)π . Then − = 2n 2 2n 4n
π/2 n (2k − 1)π π cos x dx = lim cos n→∞ 4n 2n 0 k=1 π π π π = lim cos + cos 3 + · · · + cos (2n − 1) n→∞ 2n 4n 4n π ⎤ 4n π ⎡ sin π ⎣ sin 2n · 4n ⎦ π π 2π = lim = lim n→∞ 2n n→∞ 4 2 sin n sin 4n 4n 1 π π 4 = · = 1. lim = 4 n→∞ n sin π 4 π 4n
74. The midpoint of the kth subinterval is k
5.5
Fundamental Theorem of Calculus
7
1.
3
10
2.
7
dx = x]3 = 7 − 3 = 4
2 2
3. −1
10
(−4) dx = −4x]2 = −40 − (−8) = −32
(2x + 3) dx = (x2 + 3x)
2 −1
= 10 − (−2) = 12
Integrals PROBLEMAS 1.3
5.1
The Indefinite Integral
1.
3 dx = 3x + C
2. 3. 4. 5. 6.
x5 dx =
1 6 x +C 6
5x1/4 dx = 4x5/4 + C dx √ = 3 x
√ 3
7.
x−1/3 dx =
x2 dx =
(1 − t−0.52 ) dt = t − √
3 2/3 x +C 2
x2/3 dx =
10w w dw =
8. 9.
(π 2 − 1) dx = (π 2 − 1)x + C
3 5/3 x +C 5
1 0.48 t +C 0.48
10w3/2 dw = 4w5/2 + C
(3x2 + 2x − 1) dx = x3 + x2 − x + C
√ 9 4 1 2 t − t − 2 dt = (2t1/2 − t − 9t−2 ) dt = t3/2 − t2 + 9t−1 + C t 3 2 √ 2 4 11. x(x2 − 2) dx = (x5/2 − 2x1/2 ) dx = x7/2 − x3/2 + C 7 3
10.
286
5.1. THE INDEFINITE INTEGRAL 12. 13.
5 2 √ +√ 3 2 s s3
ds =
(4x + 1)2 dx = √
2
( x − 1) dx =
14.
3
(4w − 1) dw =
15. 16.
287
(5s−2/3 + 2s−3/2 ) ds = 15s1/3 − 4s−1/2 + C
(16x2 + 8x + 1) dx = (x − 2x1/2 + 1) dx =
16 3 x + 4x2 + x + C 3 1 2 4 3/2 x − x +x+C 2 3
(64w3 − 48w2 + 12w − 1) dw = 16w4 − 16w3 + 6w2 − w + C
(5u − 1)(3u2 + 2) du =
(15u3 − 3u2 + 10u − 2) du =
15 4 u − u3 + 5u2 − 2u + C 4
r2 − 10r + 4 dr = (r−1 − 10r−2 + 4r−3 ) dr = ln |r| + 10r−1 − 2r−2 + C r3 (x + 1)2 2 4 √ 18. dx = (x3/2 + 2x1/2 + x−1/2 ) dx = x5/2 + x3/2 + 2x1/2 + C 5 3 x −1 1 1 1 x − x−2 + x−3 dx = (x−3 − x−4 + x−5 ) dx = − x−2 + x−3 − x−4 + C 19. 2 x 2 3 4 3 t − 8t + 1 1 1 1 −3 −1 −3 −4 −2 (t t 20. dt = − 8t + t ) dt = − +C ln |t| + 4t (2t)4 16 16 3 1 ln |t| 1 −2 + t − t−3 + C = 16 4 48 21. (4 sin x − 1 + 8x−5 ) dx = −4 cos x − x − 2x−4 + C 17.
22.
(−3 cos x + 4 sec2 x) dx = −3 sin x + 4 tan x + C
csc x(csc x − cot x) dx =
23. 24.
sin t dt = cos2 t
(csc2 x − csc x cot x) dx = − cot x + csc x + C
tan t sec t dt = sec t + C
2 + 3 sin2 x dx = (2 csc2 x + 3) dx = −2 cot x + 3x + C sin2 x 2 26. 40 − dθ = (40 − 2 cos θ) dθ = 40θ − 2 sin θ + C sec θ 27. (8x + 1 − 9ex ) dx = 4x2 + x − 9ex + C 25.
28.
(15x−1 − 4 sinh x) dx = 15 ln x − 4 cosh x + C
CHAPTER 5. INTEGRALS
288
2x3 − x2 + 2x + 4 5 29. dx = 2x − 1 + 2 dx = x2 − x + 5 tan−1 x + C 1 + x2 x +1 1 1 x6 1 4 2 dx = − x −x +1− 2 30. dx = x5 − x3 + x − tan−1 x + C 1 + x2 x +1 5 3 31. tan2 x dx = (sec2 x − 1) dx = tan x − x + C
32. 33.
34.
35.
36.
37.
38.
39.
cos2
x dx = 2
1 (1 + cos x) dx = 2
1 cos x + 2 2
dx =
sin x 1 x+ +C 2 2
d √ 2 1 ( 2x + 1 + C) = √ =√ dx 2 2x + 1 2x + 1
d 1 1 (2x2 − 4x)10 + C = (2x2 − 4x)9 (4x − 4) = (2x2 − 4x)9 (x − 1) dx 40 4 d 1 4 sin 4x + C = cos 4x = cos 4x dx 4 4 d 1 2 sin2 x + C = sin x cos x = sin x cos x dx 2 2 d 2x 1 sin x2 = x sin x2 − cos x2 + C = dx 2 2 (2 sin2 x) · 0 − 1 · (4 sin x cos x) cos x d 1 +C =− = − 2 dx 2 sin x 4 sin4 x sin3 x 1 d (x ln x − x + C) = x + ln x − 1 = ln x dx x
d (xex − ex + C) = xex + ex − ex = xex dx d d 1 3 (x2 − 4x + 5) dx = x − 2x2 + 5x + C = x2 − 4x + 5 41. dx dx 3 d 2 42. (x − 4x + 5) dx = (2x − 4) dx = x2 − 4x + C dx 43. y = (6x2 + 9) dx = 2x3 + 9x + C 40.
44. y = 45. y =
(10x + 3x1/2 ) dx = 5x2 + 2x3/2 + C x−2 dx = −x−1 + C = −
1 +C x
5.1. THE INDEFINITE INTEGRAL 46. y = 47. y = 48. y =
(2 + x)2 = x5
289
4 1 (4x−5 + 4x−4 + x−3 ) dx = −x−4 − x−3 − x−2 + C 3 2
(1 − 2x + sin x) dx = x − x2 − cos x + C sec2 x dx = tan x + C
49. We have f (x) = (2x − 1) dx = x2 − x + C. Solving 3 = f (2) = 4 − 2 + C = 2 + C we obtain C = 1. Thus f (x) = x2 − x + 1. √ 50. We have f (x) = x−1/2√dx = 2x1/2 + C. Solving 1 = f (9) = 2 9 + C = 6 + C we obtain C = −5. Thus f (x) = 2 x − 5. 1 2 51. f (x) = 2x dx = x + C1 ; f (x) = (x2 + C1 ) dx = x3 + C1 x + C2 3 52. We have f (x) = 6 dx = 6x + C. Solving 2 = f (−1) = −6 + C we obtain C = 8. Then f (x) = 6x+8 and f (x) = (6x+8) dx = 3x2 +8x+C. Solving 0 = f (−1) = 3−8+C = −5+C we obtain C = 5. Thus f (x) = 3x2 + 8x + 5. 53. We have f (x) = (12x2 + 2) dx = 4x3 + 2x + C. Solving 3 = f (1) = 6 + C we obtain C = −3. Then f (x) = 4x3 + 2x − 3 and f (x) = (4x3 + 2x − 3) dx = x4 + x2 − 3x + C. Solving 1 = f (1) = −1 + C we obtain C = 2. Thus f (x) = x4 + x2 − 3x + 2. 54. f (x) = an−1 xn−1 + an−2 xn−2 + · · · + a1 x + a0 55. G is an antiderivative of f . In other words, since G (x) = f (x), f is the slope function for G. Observe where G is increasing, and the graph of f is always positive. Also, G appears to have no relative extrema on the interval shown, and correspondingly the graph of f does not cross the x-axis. 56. F is an antiderivative of f . In other words, since F (x) = f (x), f is the slope function for F . Observe where the tangent lines to the graph of F have positive (negative) slope, the graph of f is positive (negative). Also, the graph of F has two relative extrema and the graph of f correspondingly crosses the x-axis. 2 ω ω2 2 57. y = x dx = x + C. From Figure 5.1.5 1.3.5 we see that y(0) = 0. Thus, 0 = y(0) = C, g 2g 2 2 ω x . and y = 2g qL L q 2 q 3 qL 2 x− x dx = x − x + C. Solving 0 = f 58. We have f (x) = = 2EI 2EI 4EI 6EI 2 qL3 qL3 qL 2 q 3 qL3 qL3 − + C we obtain C = − . Then f (x) = x − x − and 16EI 48EI 4EI 6EI 24EI 24EI 3 3 qL 2 q 3 qL q qL qL 3 f (x) = x − x − x − x4 − x + C. Solving 0 = dx = 4EI 6EI 24EI 12EI 24EI 24EI q f (0) = C we obtain C = 0. Thus f (x) = (2Lx3 − x4 − L3 x). 24EI
CHAPTER 5. INTEGRALS
290 1 d (ln | ln x| + C) = 59. dx ln x
1 1 = x x ln x
d 2 x (x e − 2xex + 2ex + C) = x2 ex + 2xex − 2xex − 2ex + 2ex = x2 ex dx 1 61. Since f (x) = x2 , f (x) = x2 dx = x3 + C. Since y = 4x + 7 is a tangent line to the 3 1 3 graph of f , then 4x + 7 = x + C at some point on f . In addition, the slope at this point 3 1 2 is 4 = f (x) = x , so x = ±2. Thus, 4(±2) + 7 = (±2)3 + C, so C = 37/3 or 5/3. Thus, 3 1 37 5 1 or f (x) = x3 + . f (x) = x3 + 3 3 3 3 60.
62. e4
4
4
= e4 ln |x|+C = eln x eC = C1 eln x = C1 x4
d 1 4 (x + 1) + C = (x + 1)3 63. dx 4 d 1 4 3 2 3 x + x + x + x + C = x3 + 3x2 + 3x + 1 = (x + 1)3 dx 4 2 Thus, both results are correct. dx/x
1 d sin πx = π cos πx, the antiderivative F of cos πx would be of the form sin πx + C. dx π 3π 1 1 1 1 + C we obtain C = . Thus, F (x) = sin πx + . Solving F (3/2) = 0 = sin π 2 π π π
64. Since
PROBLEMAS 1.4 5.2 Integration by the u-Substitution
1 (1 − 4x)1/2 (−4 dx) 1. 1 − 4x dx = − u = 1 − 4x, du = −4 dx 4 1 1 1 u1/2 du = u3/2 + C = − (1 − 4x)3/2 + C =− 4 6 6 1 (8x + 2)1/3 (8 dx) 2. (8x + 2)1/3 dx = u = 8x + 2, du = 8 dx 8 1 3 4/3 3 u1/3 du = u +C = (8x + 2)4/3 + C = 8 32 32 1 1 (5x + 1)−3 (5 dx) 3. dx = u = 5x + 1, du = 5 dx (5x + 1)3 5 1 1 1 u−3 du = − u−2 + C = − = +C 5 10 10(5x + 1)2 4. (7 − x)49 dx = − (7 − x)49 (−dx) u = 7 − x, du = −dx 1 1 = − u49 du = − u50 + C = − (7 − x)50 + C 50 50
√
5.2. INTEGRATION BY THE u-SUBSTITUTION 5.
6.
7.
8.
9.
10.
11. 12. 13.
14. 15. 16.
1 2 2 x x + 4 dx = x + 4 (2x dx) u = x2 + 4, du = 2x dx 2 1 1 1 = u1/2 du = u3/2 + C = (x2 + 4)3/2 + C 2 3 3 1 t √ (t2 + 9)−1/3 (2t dt) u = t2 + 9, dt = 2t dt dt = 3 2 2 t +9 1 3 3 u−1/3 du = u2/3 + C = (t2 + 9)2/3 + C = 2 4 4 1 (sin5 3x)(3 cos 3x dx) u = sin 3x, du = 3 cos 3x dx sin5 3x cos 3x dx = 3 1 1 6 1 u5 du = u +C = sin6 3x + C = 3 18 18 1 (cos4 2θ)(−2 sin 2θ dθ) u = cos 2θ, du = −2 sin 2θ dθ sin 2θ cos4 2θ dθ = − 2 1 1 1 u4 du = − u5 + C = − cos5 2θ + C =− 2 10 10 1 (tan2 2x)(2 sec2 2x dx) u = tan 2x, du = 2 sec2 2x dx tan2 2x sec2 2x dx = 2 1 1 1 u2 du = u3 + C = tan3 2x + C = 2 6 6 √ tan x sec2 x dx = (tan x)1/2 (sec2 x dx) u = tan x, du = sec2 x dx 2 2 = u1/2 du = u3/2 + C = (tan x)3/2 + C 3 3 1 1 (sin 4x)(4 dx) = − cos 4x + C sin 4x dx = 4 4 x x dx x 5 cos dx = 10 cos = 10 sin + C 2 2 2 2 √ √ √ 2 2 3/2 1 1 1/2 ( 2t − cos 6t) dt = 2 t dt − (cos 6t)(6 dt) = t − sin 6t + C 6 3 6 1 1 3/2 = (2t) − sin 6t + C 3 6 1 1 sin(2 − 3x) dx = − sin(2 − 3x)(−3 dx) = cos(2 − 3x) + C 3 3 1 1 (sin x2 )(2x dx) = − cos x2 + C x sin x2 dx = 2 2 cos(1/x) dx = − [cos(1/x)](−dx/x2 ) = − sin(1/x) + C x2
291
CHAPTER 5. INTEGRALS
292 17. 18. 19.
1 x sec x dx = 3 2
2
3
csc2 (0.1x) dx = csc
1 0.1
tan 5v sec 5v dv =
21.
22.
23. 24. 25. 26. 27.
28.
29.
(sec2 x3 )(3x2 dx) =
1 tan x3 + C 3
(csc2 0.1x)(0.1 dx) = −10 cot(0.1x) + C
√ √ √ √ √ √ x cot x csc x cot x √ √ dx = 2 dx u = x, du = dx/2 x x 2 x √ = 2 csc u cot u du = −2 csc u + C = −2 csc x + C
20.
1 sec 5v + C 5
1 1 1 dx = (7 dx) u = 7x + 3, du = 7 dx 7x + 3 7 7x + 3 1 1 1 1 du = ln |u| + C = ln |7x + 3| + C = 7 u 7 7 1 1 (5x + 6)−1 dx = (5 dx) u = 5x + 6, du = 5 dx 5 5x + 6 1 1 1 1 du = ln |u| + C = ln |5x + 6| + C = 5 u 5 5 1 2x dx 1 x dx = = ln(x2 + 1) + C x2 + 1 2 x2 + 1 2 1 15x2 dx 1 x2 dx = = ln |5x3 + 8| + C 3 5x + 8 15 5x3 + 8 15 x+1−1 dx x dx = dx = dx − = x − ln |x + 1| + C x+1 x+1 x+1 2 (x + 3)2 x + 6x + 9 1 1 dx = dx = x+4+ dx = x2 + 4x + ln |x + 2| + C x+2 x+2 x+2 2 1 1 1 1 dx = dx u = ln x, du = dx x ln x ln x x x 1 du = ln |u| + C = ln | ln x| + C = u 1 1 − sin θ dθ = [(1 − sin θ) dθ] u = θ + cos θ, du = (1 − sin θ) dθ θ + cos θ θ + cos θ 1 du = ln |u| + C = ln |θ + cos θ| + C = u 1 sin(ln x) 1 dx = sin(ln x) dx u = ln x, du = dx x x x = sin u du = − cos u + C = − cos(ln x) + C
5.2. INTEGRATION BY THE u-SUBSTITUTION 30.
31. 32. 33. 34. 35. 36. 37.
293
1 1 1 dx u = ln x, du = dx (ln x)2 x x 1 1 1 +C = du = − + C = − u2 u ln x 1 1 10x e10x (10 dx) = e e10x dx = +C 10 10 1 1 1 e−4x (−4 dx) = − e−4x + C dx = − 4x e 4 4 3 3 3 1 1 e−2x (−6x2 dx) = − e−2x + C x2 e−2x dx = − 6 6 3 −3 3 e1/x 1 1 ex (−3x−4 dx) = − e1/x + C dx = − x4 3 3 √ √ √ e− x 1 − x √ dx = −2 e − √ dx = −2e− x + C x 2 x √ ex dx = ex/2 dx = 2ex/2 + C 1 dx = x(ln x)2
ex − e−x dx = ex + e−x
ex
1 [(ex − e−x ) dx] + e−x
u = ex + e−x , du = (ex − e−x ) dx
1 du = ln |u| + C = ln(ex + e−x ) + C u 1 3x 3x (1 + 2e3x )1/2 (6e3x dx) 1 + 2e dx = u = 1 + 2e3x , du = 6e3x dx 38. e 6 1 1 1 u1/2 du = u3/2 + C = = (1 + 2e3x )3 + C 6 9 9 x 1 √ dx = sin−1 √ + C 39. 2 5 5−x 1 4x 1 1 1 √ 40. +C dx = (4 dx) = sin−1 2 2 4 4 3 9 − 16x 9 − (4x) 1 1 1 1 41. dx = (5 dx) = tan−1 5x + C 1 + 25x2 5 1 + (5x)2 5 1 1 x 3x 1 1 1 1 −1 42. tan + C = √ tan−1 √ + C dx = dx = 2 + 9x2 9 2/9 + x2 9 3 2 2 2/9 2/9 ex 1 43. dx = (ex dx) u = ex , du = ex dx 1 + e2x 1 + (ex )2 1 = du = tan−1 u + C = tan−1 ex + C 1 + u2 =
CHAPTER 5. INTEGRALS
294
1 1 1 θ √ dθ = (2θ dθ) = sin−1 θ2 + C 44. 4 2 2 2 2 1−θ 1 − (θ ) 2x − 3 1 3 √ √ √ 45. dx = (2x dx) − dx u = x2 , du = 2x dx 2 2 1−x 1−x 1 − x2 1 1 √ = dx = −2(1 − u)1/2 − 3 sin−1 x + C du − 3 √ 1−u 1 − x2
= −2(1 − x2 )1/2 − 3 sin−1 x + C x−8 1 1 8 46. dx = (2x dx) − dx u = x2 + 2, du = 2x dx 2 2 2 x +2 2 x +2 x +2 1 1 1 1 1 −1 x √ du − 8 +C = dx = ln |u| − 8 √ tan √ 2 u 2 ( 2)2 + x2 2 2 √ 1 x = ln(x2 + 2) − 4 2 tan−1 √ + C 2 2 47.
tan−1 x dx = 1 + x2
(tan
= 48.
u du =
sin−1 x dx = 1 − x2 =
49.
tan 5x dx =
1 5
x)
1 dx 1 + x2
u = tan−1 x, du =
(sin−1 x)1/2 u1/2 du =
√
1 dx 1 − x2
u = sin−1 x, du = √
2 3/2 2 u + C = (sin−1 x)3/2 + C 3 3
1 (tan 5x)(5 dx) = − ln | cos 5x| + C 5 (cot ex )(ex dx)
u = ex , du = ex dx
=
cot u du = ln | sin u| + C = ln | sin ex | + C
1 sin 2x + C 2 1 1 1 (1 + cos 2πx) dx = sin 2πx + C 52. cos2 πx dx = x+ 2 2 2π 1 1 1 2 (1 + cos 8x) dx = 53. cos 4x dx = x + sin 8x + C 2 2 8 1 1 1 2 3 (1 − cos 3x) dx = 54. sin x dx = x − sin 3x + C 2 2 2 3
51.
sin2 x dx =
1 dx 1 + x2
1 2 1 u + C = (tan−1 x)2 + C 2 2
ex cot ex dx =
50.
−1
1 1 (1 − cos 2x) dx = 2 2
x−
1 dx 1 − x2
5.2. INTEGRATION BY THE u-SUBSTITUTION
55.
56.
57. 58. 59.
60.
61.
62.
295
1 (1 − cos 2x) dx (9 − 12 sin x + 4 sin x) dx = 9x + 12 cos x + 4 2 1 = 9x + 12 cos x + 2 x − sin 2x + C = 11x + 12 cos x − sin 2x + C 2 1 2 2 (1 + cos 4x) dx (1 + cos 2x) dx = (1 + 2 cos 2x + cos 2x) dx = x + sin 2x + 2 1 3 1 1 = x + sin 2x + x + sin 4x + C = x + sin 2x + sin 4x + C 2 4 2 8 √ 3 3 y= 1 − x dx = − (1 − x)1/3 (−dx) = − (1 − x)4/3 + C 4 (1 − tan x)5 1 y= dx = − (1 − tan x)5 (− sec2 x dx) = − (1 − tan x)6 + C 2 cos x 6 We have f (x) = (1 − 6 sin 3x) dx = x + 2 cos 3x + C. Solving −1 = f (π) = π + 2 cos 3π + C = π − 2 + C we obtain C = 1 − π. Thus f (x) = x + 2 cos 3x + 1 − π. 1 1 (1 + 2x)6 + C. Solving 0 = f (0) = + C we obtain We have f (x) = (1 + 2x)5 dx = 12 12 1 C = − . Then 12
1 1 1 1 1 6 6 7 (1 + 2x) − [(1 + 2x) − 1] dx = (1 + 2x) − x + C. f (x) = dx = 12 12 12 12 14 1 1 1 1 1 1 . Thus f (x) = (1+2x)7 − x− . Solving 0 = f (0) = +C we obtain C = − 12 14 168 168 12 168 (a) sin x cos x dx = sin x(cos x dx) u = sin x, du = cos x dx 1 1 = u du = u2 + C1 = sin2 x + C1 2 2 (b) sin x cos x dx = − cos x(− sin x dx) u = cos x, du = − sin x dx 1 1 = − u du = − u2 + C2 = − cos2 x + C2 2 2 1 1 sin 2x dx = − cos 2x + C3 (c) sin x cos x dx = 2 4 d 1 sin2 x + C1 = sin x cos x (a) dx 2 d 1 2 − cos x + C2 = cos x sin x dx 2 d 1 1 − cos 2x + C3 = sin 2x = sin x cos x dx 4 2 2
(3 − 2 sin x) dx =
2
296
63.
64.
65.
66.
67.
CHAPTER 5. INTEGRALS 1 1 1 1 1 2 2 2 (b) sin x + C1 = (1 − cos x) + C1 = − cos x + C1 + = − cos2 x + C2 2 2 2 2 2 1 1 2 2 (c) sin x cos x dx + sin x cos x dx = sin x + C1 − cos x + C2 2 2 1 2 sin x cos x dx = − (cos2 x − sin2 x) + (C1 + C2 ) 2 1 1 sin x cos x dx = − (cos2 x − sin2 x) + (C1 + C2 ) 4 2 1 = − cos 2x + C3 4 s L sin−1 (a) From the given derivative, we have t(s) = + C. Solving t(0) = 0, we g sC obtain C = 0. sC L L π L −1 −1 (b) t(sC ) = sin sin 1 = = g sC g 2 g π L L . (c) By symmetry, T = 4t(sC ) = 4 = 2π 2 g g y = cos3 x dx = cos2 x cos x dx = (1 − sin2 x) cos x dx = cos x − (sin2 x)(cos x dx) u = sin x, du = cos x dx 1 1 = cos x − u2 du = sin x − u3 + C = sin x − sin3 x + C 3 3 2 2 1 2 1 Solving f (π/2) = 0 = 1− +C = +C we obtain C = − . Thus f (x) = sin x− sin3 x− . 3 3 3 3 3
2 1 1 (1 + cos 2x) dx = (1 + 2 cos 2x + cos2 2x) dx cos4 x dx = (cos2 x)2 dx = 2 4 1 1 1 1 1 1 1 = x + sin 2x + (1 + cos 4x) dx = x + sin 2x + (1 + cos 4x) dx 4 4 4 2 4 4 8 1 1 1 3 1 1 1 sin 4x + C = x + sin 2x + sin 4x + C = x + sin 2x + x + 4 4 8 32 8 4 32
2 1 1 4 2 2 (1 − cos 2x) dx = (1 − 2 cos 2x + cos2 2x) dx sin x dx = (sin x) dx = 2 4 1 1 1 1 1 1 1 = x − sin 2x + (1 + cos 4x) dx = x − sin 2x + (1 + cos 4x) dx 4 4 4 2 4 4 8 1 1 1 3 1 1 1 sin 4x + C = x − sin 2x + sin 4x + C = x − sin 2x + x + 4 4 8 32 8 4 32 1 1 √ √ u = x2 , du = 2x dx dx = (2x dx) 4 2 x x − 16 2x x4 − 16 u 1 1 x2 1 1 √ +C = du = sec−1 + C = sec−1 2 4 4 4 4 u u 2 − 42
5.2. INTEGRATION BY THE u-SUBSTITUTION 68.
69.
297
ex e − x dx e +1 1 (ex dx) u = ex + 1, du = ex dx = ex dx − ex + 1 1 du = ex − ln |u| + C = ex − ln(ex + 1) + C = ex − u
e2x dx = ex + 1
x
1 1 + cos x 1 + cos x 1 + cos x dx = dx dx = 1 − cos x 1 + cos x 1 − cos2 x sin2 x
cos x 1 + dx = (csc2 x + csc x cot x) dx = sin2 x (sin x)(sin x) = − cot x − csc x + C
1 dx = 1 − cos x
1 1 − sin 2x 1 − sin 2x 1 − sin 2x 70. dx dx = dx = 2 1 + sin 2x 1 − sin 2x cos2 2x 1 − sin 2x
1 sin 2x − = dx = (sec2 2x − sec 2x tan 2x) dx 2 cos 2x (cos 2x)(cos 2x) 1 1 = tan 2x − sec 2x + C 2 2 1 71. f (8x) dx = f (8x)(8 dx) u = 8x, du = 8 dx 8 1 1 1 f (u) du = f (u) + C = f (8x) + C = 8 8 8
72.
73.
1 f (5x2 )(10x dx) xf (5x ) dx = u = 5x2 , du = 10x dx 10 1 1 1 f (u) du = f (u) + C = f (5x2 ) + C = 10 10 10
2
1 [f (2x)]1/2 [2f (2x) dx] f (2x)f (2x) dx = u = f (2x), du = 2f (2x) dx 2 1 1 1 u1/2 du = u3/2 + C = [f (2x)]3/2 + C = 2 3 3
74.
1 dx = 1 + sin 2x
f (3x + 1) 1 1 dx = [3f (3x + 1) dx] u = f (3x + 1), du = 3f (3x + 1) dx f (3x + 1) 3 f (3x + 1) 1 1 1 1 du = ln |u| + C = ln |f (3x + 1)| + C = 3 u 3 3
75. For any f ,
1 f (4x)(4 dx) u = 4x, du = 4 dx 4 1 1 1 f (u) du = f (u) + C = f (4x) + C = 4 4 4
f (4x) dx =
CHAPTER 5. INTEGRALS
298 Given f (x) =
√
x4 + 1 = (x4 + 1)1/2 , we have f (x) = 2x3 (x4 + 1)−1/2 . Thus,
f (4x) dx =
1 1 32x3 f (4x) + C = {2(4x)3 [(4x)4 + 1]−1/2 } + C = √ + C. 4 4 256x4 + 1
16384x6 96x2 , − 256x4 + 1 (256x4 + 1)3 6x2 4x6 , we have f (4x) = − which should be the same as f (4x). Since f (x) = √ x4 + 1 (x4 + 1)3 6(4x)2 4(4x)6 96x2 16384x6 − =√ . − 256x4 + 1 (4x)4 + 1 [(4x)4 + 1]3 (256x4 + 1)3 76. First evaluating sec2 3x dx, we get To check this, take the derivative of the above function, yielding √
1 (sec2 3x)(3 dx) u = 3x, du = 3 dx 3 1 1 1 sec2 u du = tan u + C = tan 3x + C = 3 3 3 1 tan 3x + C dx, we get Next, evaluating sec2 3x dx dx = 3 1 1 tan 3x + C dx = (Cx + C1 ) + tan 3x dx 3 3 1 = (Cx + C1 ) + (tan 3x)(3 dx) u = 3x, du = 3 dx 9 1 1 tan u du = (Cx + C1 ) − ln | cos u| + C2 = (Cx + C1 ) + 9 9 1 = Cx − ln | cos 3x| + C3 . 9
5.3
sec2 3x dx =
The Area Problem
1. 3 + 6 + 9 + 12 + 15 2. −1 + 1 + 3 + 5 + 7 3. 2 + 2 + 8/3 + 4 4.
9 27 81 3 + + + 10 100 1000 10, 000
1 1 1 1 1 1 1 1 1 1 + − + − + − + 5. − + − 7 9 11 13 15 17 19 21 23 25 6. 1 −
1 1 1 1 1 1 1 1 1 + − + − + − + − 4 9 16 25 36 49 64 81 100
π/2 0
2 2n 4n
(2k − 1)π π cos x dx = lim cos n→∞ 4n 2n k=1 π π π π cos + cos 3 + · · · + cos (2n − 1) = lim n→∞ 2n 4n 4n π ⎤ 4n π ⎡ sin π ⎣ sin 2n · 4n ⎦ π π 2π = lim = lim n→∞ 2n n→∞ 4 2 sin n sin 4n 4n 1 π 4 π = · = 1. lim = 4 n→∞ n sin π 4 π 4n n
2n
PROBLEMAS 1.5
5.5
Fundamental Theorem of Calculus
7
1.
3
10
2.
7
dx = x]3 = 7 − 3 = 4
2 2
3. −1
10
(−4) dx = −4x]2 = −40 − (−8) = −32
(2x + 3) dx = (x2 + 3x)
2 −1
= 10 − (−2) = 12
5.5. FUNDAMENTAL THEOREM OF CALCULUS
4
t2 dt =
4. −5
3
5.
1 1
−2
= −5
π/2
π/2
sin x dx = − cos x]0
0
8.
cos θdθ =
π/4 sin θ]−π/3
cos 3t dt =
1 sin 3t 3
−π/3 π/2
9. π/4
10.
3/4 1/2
−1
12. −3
1 −1
2
14. 0 2
1 1 du = − u2 u
3/4 1/2
1
= 51 − 5 = 46
= −34 − 200 = −234
π/2 = π/4
√ √ √ 3 2+ 3 − = 2 2
√ √ 2 3π 1 3π 1 2+ 2 1 sin − sin =− − =− 3 2 3 4 3 6 6
1 1/2
=−
1 1 (cos 2π − cos π) = − 2π π
2 4 = − − (−2) = 3 3
2 −1 dx = 2 ln |x|]−3 = 0 − 2 ln 3 = − ln 9 x 1
13.
2 − = 2
1 cos 2πx sin 2πx dx = − 2π 1/2
−2
3
= 0 − (−1) = 1
1
11.
1
√
π/4
1 [64 − (−125)] = 63 3
(12x5 − 36) dx = (2x6 − 36x)
7.
4
(6x2 − 4x + 5) dx = (2x3 − 2x2 + 5x)
6.
1 3 t 3
319
ex dx = ex ]−1 = e −
1 e
(2x − 3ex ) dx = x2 − 3ex
2
2 0
= (4 − 3e2 ) − (−3) = 7 − 3e2
1 2 1 3 x − x 2 3
2
2 8 −0=− 3 3 0 0 0 3 2 3 1 3 25 16. x − 2x2 x(x − 2)(x + 2) dx = − (x3 − 4x) dx = − =− 4 4 3 2 2 1 1 7 4 2 3 5 2 17. x − x + x − 4x (7x3 − 2x2 + 5x − 4) dx = 4 3 2 −1 −1 7 2 5 7 2 5 28 − + −4 − + + +4 =− = 4 3 2 4 3 2 3 −1 −1 1 3 122 1 18. x − 2x2 + 8x (x2 − 4x + 8) dx = = − − 2 − 8 − (−9 − 18 − 24) = 3 3 3 −3 −3
15.
x(1 − x) dx =
(x − x2 ) dx =
=2−
CHAPTER 5. INTEGRALS
320
4
19. 1
4
20. 2 √
x−1 √ dx = x
3
1 1/4
22. 0
12
23.
4 1
x2 + 8 dx = x2
21.
4 2
(1 + 8x−2 ) dx = (x − 8x−1 )
1 1 √ dx = 2 2 1 − 4x
√
z + 4 dz
1/4 0
16
(2x + 1)−1/3 dx =
3 0
√
x dx + 16
1 −2
t dt (t2 + 1)2
1 2
27. 1/2
4
28. 1
1 1+ x
3
2
1 2
25 16
2 5
1
x−
8 x
4 2
= 2 − (−2) = 4
2
dx =
1 − (2x)2
π 1 −1 1/4 1 π sin 2x 0 −0 = = 2 2 6 12
2 3/2 u 3
16 = 0
128 3
8
u−1/3 du =
1
1 2
3 2/3 u 2
8 1
=3−
9 3 = 4 4
1 1 √ du = 2 u
25
u−1/2 du =
16
√ 25 u 16 = 5 − 4 = 1
u = t2 + 1, du = 2t dt =
− 2
u = x2 + 16, du = 2x dx
x2
=
26.
1
2 16 8 −4 − −2 = 3 3 3
u = 2x + 1, du = 2 dx
0
25.
u1/2 du =
0
=
u = z + 4, du = dz
7/2
4
4
√
=
24.
2 3/2 x − 2x1/2 3
3 1 π π π dx = tan−1 x 1 = − = 2 1+x 3 4 12
−4
(x1/2 − x−1/2 ) dx =
1 dx x2
√ 1+4 x √ dx x
1/2 1 du = − 2 u 2
5
u 2
−2
1 du = u−1 2
5 2
1 = 2
1 1 − 5 2
1 1 , du = − 2 dx x x
3 2 3 65 1 4 81 3 3 −4= =− u du = u du = u = 4 4 4 3 2 2 u=1+
√ 2 u = 1 + 4 x, du = √ dx x 9 9 1 1 3 4/3 3 1/3 u = u du = = (94/3 − 54/3 ) 2 5 2 4 8 5
=−
3 20
5.5. FUNDAMENTAL THEOREM OF CALCULUS
1
29. 0
√
x+1 dx + 2x + 3
u = x2 + 2x + 3, du = 2(x + 1) dx
x2
=
1
1 2
π/8
sec2 2x dx 1 2
=
32.
√
1 4
π/4
33. −1/2
4
34. 1
0
x csc x2 cot x2 dx
(x − cos πx) dx =
√ cos x √ dx 2 x
u=
2 1
π/2
35.
√
√
π/4 = 0
1 2
π/2 π/4
1 csc u cot u du = − csc u 2
1 2 1 x − sin πx 2 π
3/2
π/4
−1/2
π/2
9 1 − − = 8 π
√ 1 = − (1 − 2) = 2
√
2−1 2
1 1 − − − =1 8 π
1 x, du = √ dx 2 x 2
u = cos x, du = − sin x dx
cos x sin x dx
=−
0
√
1
u du =
1
u1/2 du =
0
2 3/2 u 3
1 = 0
2 3
π/3
sin x cos x dx
36.
u = sin x, du = cos x dx
π/6
√
3/2
= 1/2
1 tan u 2
cos u du = sin u]1 = sin 2 − sin 1
0
3
√ √ 6 √ u 3 = 6− 3
u = x2 , du = 2x dx
=
u−1/2 du =
1 dz = 0 z5
sec2 u du =
π/4
3/2
4 4
1 = 2
6
u = 2x, du = 2 dx
0
√π/2
z = u4 + 2u2 + 1, du = 4(u3 + u) du =
31.
1 1 √ du = 2 u
3
u +u du (u4 + 2u2 + 1)5
−1
6
3
30.
321
π/2
37. π/6
1 + cos θ dθ (θ + sin θ)2
1 u du = u2 2
√3/2 = 1/2
1 3 1 − = 8 8 4
u = θ + sin θ, du = (1 + cos θ) dθ
(π+2)/2
= (π+3)/6
u−2 du = −
1 u
(π+2)/2 (π+3)/6
=−
6 4π + 6 2 + = π+2 π+3 (π + 3)(π + 2)
CHAPTER 5. INTEGRALS
322
π/4
(sec x + tan x)2 dx =
38. −π/4
π/4 −π/4 π/4
=
(sec2 x + 2 sec x tan x + tan2 x) dx
−π/4 π/4
= −π/4
(sec2 x + 2 sec x tan x + sec2 x − 1) dx (2 sec2 x + 2 sec x tan x − 1) dx π/4
= (2 tan x + 2 sec x − x)]−π/4 √ √ π π 8 − π − −2 + 2 2 + = = 2+2 2− 4 4 2
3/4
39.
3/4
1 (1 − cos 2πx) dx = 2 0 1 3π 3 1 3 sin = + = − 8 4π 2 8 4π
sin2 πx dx =
0
1 1 x− sin 2πx 2 4π
3/4 0
40. Using the fact that f (x) = cos2 x is even, we have
π/2
cos2 x dx = 2
−π/2
=
5
41. 1
1 dx 1 + 2x
π/2
cos2 x dx = 2
0
x+
1 sin 2x 2
π/2 0
π/2
1 (1 + cos 2x) dx 2 0 π π +0 −0= . = 2 2
u = 1 + 2x, du = 2 dx =
1 2
11 3
1 1 du = ln |u| u 2
11 = 3
1 (ln 11 − ln 3) 2
1
42. Since f (x) = tan x is an odd function on [−1, 1], we have
tan x dx = 0. −1
43.
44.
45.
d dx d dx d dt
d 46. dx
x
tet dt = xex
0
x
ln t dt = ln x 1
t 2
(3x2 − 2x)6 dx = (3t2 − 2t)6
9 x
3
u2 + 2 du = −
3
x2 + 2
5.5. FUNDAMENTAL THEOREM OF CALCULUS 47.
d dx
6x−1
√
4t + 9 dt
3
√
u = 6x − 1, du = 6 dx u √ du √ d du = 4u + 9 = 4t + 9 dt du dx dx 3 √ = 4(6x − 1) + 9 · 6 = 6 24x + 5
√ 1 u = x = x1/2 , du = x−1/2 dx 2 π u 1 −1/2 du d 1 2 2 du = sin u = sin x · x = sin t dt = √ sin x du dx dx 2 2 x π x2 x2 0 3x d 1 1 1 1 d 49. F (x) = dt + dt = − dt + dt 3 dx t3 + 1 dx t3 + 1 t3 + 1 3x t + 1 0 0 0
d 48. dx
x
sin t2 dt
u = 3x, du = 3 dx; z = x2 , dz = 2x dx u z du dz d 1 1 d dt + dt = − 3 3 du dx dz dx 0 t +1 0 t +1 1 2x 3 1 (3) + 2 3 (2x) = 6 − =− (3x)3 + 1 (x ) + 1 x + 1 27x3 + 1 0 5x d 2 2 50. F (x) = t + 1 dt + t + 1 dt dx x 0 sin sin x 5x d − = t2 + 1 dt + t2 + 1 dt dx 0 0 u = sin x, du = cos x dx; z = 5x, dz = 5 dx u z du dz d d 2 2 + = t + 1 dt t + 1 dt − du dx dz dx 0 0 2 2 2 = − sin x + 1(cos x) + (5x) + 1(5) = 5 25x + 1 − cos x sin2 x + 1
x d (2t3 − 4t2 + 5t) 1 dx 1 d [(2x3 − 4x2 + 5x) − (2 − 4 + 5)] = 6x2 − 8x + 5 = dx
d π x d d t x t t t sin dx = = 52. −3 cos −3 cos − −3 cos = sin dt π 3 dt 3 π dt 3 3 3
51.
d dx
x
(6t2 − 8t + 5) dt =
1
53. (a) f (1) =
ln(2t + 1) dt = 0 1
(b) f (x) = ln(2x + 1), so f (1) = ln[2(1) + 1] = ln 3. 2 2 2 , so f (1) = = . (c) f (x) = 2x + 1 2(1) + 1 3
323
CHAPTER 5. INTEGRALS
324 4 4 4 , so f (1) = − =− . (2x + 1)2 [2(1) + 1]2 9
(d) f (x) = − 54. (a) G(x2 ) =
x2
f (t) dt a
x2 d d 2 G(x ) = f (t) dt = 2xf (x2 ) dx dx a x3 +2x (c) G(x3 + 2x) = f (t) dt
(b)
a
2
55.
2
f (x) dx +
19 1 1 =− 0− + (8 − 0) = 2 3 6 0 2 f (x) dx = f (x) dx + f (x) dx =
−1
−1
= (x2 + 3x)
3
0
0 −1
f (x) dx =
0
1 x dx = − x2 2 2
0
0
1 + x3 3 −1
2 0
2
(2x + 3) dx +
3 dx
−1
0
2
0
3
f (x) dx +
0
−x dx +
−1
2
+ 3x]0 = [0 − (−2)] + 6 = 8
2
0
f (x) dx = 0
56.
57.
f (t) dt = (3x2 + 2)f (x3 + 2x)
a
−1
2
x3 +2x
0
f (x) dx = −1
d d G(x3 + 2x) = dx dx
(d)
2
f (x) dx = 2
3
4 dx + 0
2
2
3
dx = 4x]0 + x]2
= (8 − 0) + (3 − 2) = 9
π
58.
π/2
f (x) dx =
π
f (x) dx +
0
0
f (x) dx =
π/2
π
sin x dx + 0
π/2
= − cos x]0
π/2
cos x dx π/2
π
+ sin x]π/2 = −(0 − 1) + (0 − 1) = 0
59. Using the fact that f (x) is an even function on [−2, 2], we have
2
f (x) dx = 2 −2
4
60. 0
x dx =
1 0
=
1 4x]0
x dx +
1
0 dx + 0
1 + x3 3 2
1 2
1
0
38 1 . = 2 (4 − 0) + (8 − 1) = 3 3
x dx + 3
3 2
x dx +
4
2 dx + 2
= (2 − 1) + (6 − 4) + (12 − 9) = 6
3
4 3
1
f (x) dx = 2 1
2
f (x) dx +
2
1 dx + 1
1
f (x) dx = 2 0
=2
2
4 dx + 0
3
4
3 dx = x]1 + 2x]2 + 3x]3
2
x dx 1
x dx 2
2
5.5. FUNDAMENTAL THEOREM OF CALCULUS
325
0
1 1 1 9 1 x dx = − x2 + x2 = + = 5 2 2 2 2 −3 −3 0 −3 0 4 3 4 3 4 62. |2x − 6| dx = −(2x − 6) dx + (2x − 6) dx = (−x2 + 6x) 0 + (x2 − 6x) 3
1
|x| dx =
61.
0
−x dx +
0
1
0
3
= (9 − 0) + [(−8) − (−9)] = 10
0
3 0 3 3 √ √ 2 2 3/2 3/2 |x| + 1 dx = −x + 1 dx + x + 1 dx = − (1 − x) + (x + 1) 63. 3 3 −8 −8 0 −8 0 2 2 = − (1 − 27) + (8 − 1) = 22 3 3 1 2 2 1 2 1 3 1 64. x −x |x2 − 1| dx = −(x2 − 1) dx + (x2 − 1) dx = − x3 + x + 3 3 0 0 1 0 1
2 2 2 −0 + − − = =2 3 3 3 65. Using the fact that f (x) = | sin x| is an even function on [−π, π] and sin x > 0 for 0 ≤ x ≤ π, π π π π | sin x| dx = 2 | sin x| dx = 2 sin x dx = −2 cos x]0 = −2(−1 − 1) = 4. −π
π
66. 0
0
| cos x| dx =
0
π/2
π
cos x dx + 0
π/2
π/2
(− cos x) dx = sin x]0
π
− sin x]π/2
= (1 − 0) + (0 − 1) = 2
e
67. 1/2
(ln 2t)5 dt t
1 dt; u(1/2) = 0, u(e) = 1 + ln 2 t
1+ln 2 1+ln 2 1 1 (1 + ln 2)6 ≈ 3.9266 = u5 du = u6 = [(1 + ln 2)6 − 0] = 6 6 6 0 0 u = ln 2t, du =
68. (Ask Scott for analytic value of arctan sqrt(2)/2, if any)
1.4.63 and 64 include definite integral notation, seemingly before it is used, (While we’re at it, 5.2.63 so ask about their placement also) (One last set of questions: 84 and [misnumbered] 86 seem to be open-ended; 86 has a few typos in addition to the numbering) (And, of course, don’t forget to ask about whatever is blank) 1 1 dx √ −1 x)(1 + x2 ) 2/2 (tan √ √ 1 π dx; u( 2/2) = tan−1 ( 2/2), u(1) = 2 1+x 4 √ 1 2 π π/4 du = ln |u|]tan−1 √2/2 = ln − ln tan−1 ≈ 0.2438 u 4 2
u = tan−1 x, du = =
π/4 √ tan−1 ( 2/2)
CHAPTER 5. INTEGRALS
326
1
69. 0
e−2x dx +1
u = e−2x + 1, du = −2e−2x ; u(0) = 2, u(1) = 1 + e−2
e−2x
1 =− 2
1/ 2
70. 0
√
1 1 du = − ln |u| u 2
2
√
1+e−2
1+e−2
1 = − [ln(1 + e−2 ) − ln 2] ≈ 0.2831 2
2
√ u = x2 , du = 2x dx; u(0) = 0, u(1/ 2) = 1/2
x dx 1 − x4 =
1 2
1/2 0
√
1 1 du = sin−1 u 2 1 − u2
1/2 = 0
π 1 π −0 = ≈ 0.2618 2 6 12
2 2 71. (a) Since erf (x) = √ e−x > 0, erf(x) is increasing for all x. π √ √ dy 2 2 (b) The derivative of y = ex [1 + π erf(x)] is = 2 + 2ex x[1 + π erf(x)], so dx √ √ 2 2 dy − 2xy = 2 + 2ex x[1 + π erf(x)] − 2xex [1 + π erf(x)] = 2 dx √ √ Also, y(0) = e0 [1 + π erf(0)] = 1 + π · 0 = 1. x sin x sin t 72. (a) Si(x) = , and so Si (x) = 0 for x = nπ, n = 1, 2, . . . . The first dt, Si (x) = x t 0 x cos x − sin x four positive critical numbers are then π, 2π, 3π, and 4π. Now, Si (x) = , x2 therefore 1 1 1 1 Si (π) = − < 0, Si (2π) = > 0, Si (3π) = − < 0, Si (4π) = >0 π 2π 3π 4π shows that there are relative maxima at x = π and x = 3π and relative minima at x = 2π and x = 4π. (b) 2
1
73.
74.
lim
P →0
lim
P →0
n
2
3
(2x∗k + 5)Δxk =
k=1 n
4
cos
k=1
x∗k Δxk = 4
5
3
6
(2x + 5) dx = (x2 + 5x)
−1 2π
cos 0
π sin x∗k = lim (sin x∗k )Δxk = lim n→∞ n n→∞ k=1
n
k=1
−1
= 24 − (−4) = 28
x 2π x dx = 4 sin =4 4 4 0
75. Letting Δxk = π/n we have n
3
π 0
π
sin x dx = − cos x]0 = −(−1 − 1) = 2.
5.5. FUNDAMENTAL THEOREM OF CALCULUS
327
76. Letting Δxk = 2/n we have 2 ∗ xk = lim x∗k Δxk = n→∞ n n→∞ n
n
1
x dx =
lim
k=1
2
x
77. −1
π/2
k=1
12t2 dt dx =
1
4t3
x
−1
t
78. 0
2
sin x dx dt =
0
π/2 0
1
2
dx = −1
−1
π/2
1 = −1
1 (1 − 1) = 0. 2
(4x3 − 4) dx = (x4 − 4x)
t − cos x]0 dt =
= (− sin t + t)]0
1 2 x 2
2 −1
=8−5=3
π/2
(− cos t + 1) dt 0
π π−2 = −1 + −0= 2 2
79. Since f (x) is even, f (−x) = f (x). Then
a −a
0
f (x) dx =
a
f (x) dx +
t = −x, dt = −dx
f (x) dx
−a 0
0
a 0 a f (−t)(−dt) + f (x) dx = − f (t) dt + f (x) dx a 0 a 0 a a a = f (t) dt + f (x) dx = 2 f (x) dx. =
0
0
0
80. (a) Since f is odd and continuous at x = 0, f (0) = 0. (b) 3
-3
3 -3
−3
(c) F (−3) =
3
f (t) dt = 0 since f is odd.
f (t) dt = 0; F (3) = −3
−3
(d) 6
3
-3
3
(e) Since F (x) = f (x), critical numbers occur at x = −3, x = 0, and x − 3. Solving F (x) = f (x) = 0 we see that points of inflection occur at x = −2 and x = 2. √ 81. The reasoning is flawed at the point that sin t is substituted with 1 − cos2 t. The use of the square root loses sin t’s sign changes.
CHAPTER 5. INTEGRALS
328 82. (a)
d x dx
2x
t3 + 7 dt
u = 2x, du = 2 dx u
d du d du d x x = t3 + 7 dt u3 + 7 = dx du dx dx dx 1 d d x 2 (2x)3 + 7 = 2x 8x3 + 7 = dx dx
1 40x3 + 14 = 2x (8x3 + 7)−1/2 (24x2 ) + 2 8x3 + 7 = √ 2 8x3 + 7 4 4 4 d (b) x t3 + 7 dt = t3 + 7 dt, since t3 + 7 dt is a constant. dx 1 1 1 1
83. (a) 1
1
2
-1
-1
f (x) = cos3 x 2π 2π 3 ; cos x dx = 0 sin3 x dx = 0
(b) 0
2
f (x) = sin3 x
0
84. As this project’s exact results may vary for every “run” of the exercise, no exact solution is given. In general, the student should see the empirical probability n/N approach the area of the region as the number of random points increases. 85. (a) At time n the radius of the circle is r0 + cu and the area is A(u) = π(r0 + cu)2 . Then
kπ(r0 + cu)2 Kπ t du = (r0 + cu)2 (c du) V cV 0 0 0 0
t Kπ 1 Kπ (r0 + ct)3 − r03 (r0 + cu)3 = = cV0 3 3cV 0 0
RT = Pv
t
3cV0 RT = (r0 + ct)3 − r03 P Kvπ 3cV0 RT + r03 (r0 + ct)3 = P Kvπ 3 3cV0 RT + r03 r0 + ct = P Kvπ 1 3 3cV0 RT r0 + r03 − . t= c P Kvπ c (b) Substituting RT /P v = 1.9 × 106 , K = 0.01 × 10−3 , c = 0.01, r0 = 100, and V0 = 10, 000, we find t ≈ 2, 617, 695 seconds, or t ≈ 30 days and 7 hours. (c) The final area is A(2, 617, 695) = π[100+0.01(2, 617, 695)]2 ≈ 2.169×109 m2 = 2169 km2 .
CHAPTER 5 IN REVIEW
329
86. Since this exercise involves a research report, no solution is given. The need for the definite θ0 sec x dx can be found in the derivation of the projection, whose key properties integral 0
are that it is conformal (i.e., angle-preserving) and that it represent lines of constant course as straight segments.
PROBLEMAS DE REPASO DE LA UNIDAD 1
Chapter 5 in Review A. True/False
1. False. Consider f (x) = x3 + x2 + 1. 2. True 3. True 4. True 5. True 6. False. Continuity implies integrability, but not necessarily the other way around. Consider the discontinuous function 0, x = 1 f (x) = 1, x = 1 which is integrable on [0, 2] by (15) in Section 5.4. 7. True, since no portion of the graph of y = x − x3 lies below the x-axis on [0, 1]. 8. False. This is only true when no portion of the graph lies below the x-axis. 1 1 1 9. False. Consider the partition 0, , , . . . , , 1 of {0, 1}. n n−1 2 10. True 11. True 12. True 13. False. 14. True 15. True 16. True
sin x dx = − cos x + C.
CHAPTER 5. INTEGRALS
330
B. Fill in the Blanks 1. f (x) 2. x2 + C ln x x √ 4. 6
3.
5. −f (g(x))g (x) dx 6.
7.
5 1 √ − 2ex x e25x2 5 k=1
k 2k + 1
8. 3480 17 9. 5
10. 4;
3
11. 5/2 12. regular
4
13.
√
4 2 3/2 16 x = 3 3 0
x dx;
0
14. −4 1 15. −1
1 −1
x
e 0
−t
dt
1
dx = −1
!
x " −e−t 0
dx = −
1 −1
(e−x − 1) dx
1 1 = −(−e−x − x) −1 = −[(−e−1 − 1) − (−e1 + 1)] = − e + 2 e x 1 d 1 1 e−t dt dx = e−x dx = −e−x −1 = e − dx e 0 −1
16. 4/3
C. Exercises
1
1. −1
9
2. 1
(4x3 − 6x2 + 2x − 1) dx = (x4 − 2x3 + x2 − x) 9 6 √ dx = 12x1/2 = 36 − 12 = 24 x 1
1 −1
= −1 − 5 = −6
CHAPTER 5 IN REVIEW
(5t + 1)100 dt =
3.
w2
4.
π/4
5. 0
331
1 (5t + 1)101 + C 505
3w3 + 1 dw
u = 3w3 + 1, du = 9w2 dw 1 √ 1 2 3/2 2 u1/2 du = u +C = = u du = (3w3 + 1)3 + C 9 9 27 27
(sin 2x − 5 cos 4x) dx =
π2
6. π 2 /9
√ sin z √ dz z
u=
√
5 1 − cos 2x − sin 4x 2 4
1 z, du = √ dz 2 z
π
=
2 sin u du = π/3
4
7.
π/4
π −2 cos u]π/3
0
1 =0− − 2
1 = −2 −1 − 2
=
1 2
=3
(−2x2 + x1/2 ) dx = 0
4
π/4
π/4
dx +
8. −π/4
tan2 x dx =
−π/4
9.
10. 11.
cot6 8x csc2 8x dx
π/4
(1 + tan2 x) dx =
−π/4
=
π/4
sec2 x dx
−π/4
π/4 tan x]−π/4
= 1 − (−1) = 2
u = cot 8x, du = −8 csc2 x dx 1 1 1 u6 du = − u7 + C = − cot7 8x + C =− 8 56 56
1 csc 3x cot 3x dx = − csc 3x + C 3 (4x2 − 16x + 7)4 (x − 2) dx =
1 8
(4x2 − 16x + 7)4 [8(x − 2) dx]
u = 4x2 − 16x + 7, du = 8(x − 2) dx 1 1 5 1 u4 du = u +C = (4x2 − 16x + 7)5 + C = 8 40 40 12.
(x2 + 2x − 10)2/3 (5x + 5) dx =
5 2
(x2 + 2x − 10)2/3 [2(x + 1) dx]
u = x2 + 2x − 10, du = 2(x + 1) dx 5 3 3 u2/3 du = u5/3 + C = (x2 + 2x − 10)5/3 + C = 2 2 2
CHAPTER 5. INTEGRALS
332 √ 3
13.
x2 + 1 1 dx = 3 3 x + 3x − 16
(x3 + 3x − 16)−1/3 [3(x2 + 1) dx]
u = x3 + 3x − 16, du = 3(x2 + 1) dx 1 1 1 u−1/3 du = u2/3 + C = (x3 + 3x − 16)2/3 + C = 3 2 2
1 x2 + 1 dx = 3 x + 3x − 16 3
14.
1 [3(x2 + 1) dx] x3 + 3x − 16
u = x3 + 3x − 16, du = 3(x2 + 1) dx 1 1 1 1 du = ln |u| + C = ln |x3 + 3x − 16| + C = 3 u 3 3
4
15. 0
x dx 16 + x2
u = x2 + 16, du = 2x dx 1 2
=
4
16. 0
2
17. 0
2
18. 0
32 16
1 x 1 dx = tan−1 16 + x2 4 4
cot 10x dx =
u = 16 − x2 , du = −2x dx 12
u−1/2 du =
16
5
4
f (x) dx =
9
f (x) dx + 1
7
f (x) dx;
0 9
√ √ √ 16 √ u 12 = 16 − 12 = 4 − 2 3
1 ln | cos 10x| + C 10
f (x) dx +
0
1
5
f (x) dx =
22.
1 2
1 ln | sin 10x| + C 10
7
1 π (tan−1 1 − tan−1 0) = 4 16
x dx 16 − x2
0
16
1 32 1 1 (ln 32 − ln 16) = ln = ln 2 2 2 16 2
√
tan 10x dx = −
21.
=
=
1 x 2 1 π dx = sin−1 = sin−1 − sin−1 0 = 4 0 2 6 16 − x2
20.
4
32
√
=−
19.
1 1 du = ln |u| u 2
4
2 = −3 +
7
7
f (x) dx; 5
f (x) dx = 2 + (−8) = −6
f (x) dx = 5 5
CHAPTER 5 IN REVIEW 23. Since |x − 1| =
3 0
−x + 1, x − 1,
333 0≤x 0 we see that w2 < . B u=1+
PROBLEMAS 2.3
7.3
Integration by Parts
1.
√ x x + 3 dx
2.
√
x dx 2x − 5
2 u = x, du = dx; dv = (x + 3)1/2 dx, v = (x + 3)3/2 3 2 2 2 4 (x + 3)3/2 dx = x(x + 3)3/2 − (x + 3)5/2 + C = x(x + 3)3/2 − 3 3 3 15 u = x, du = dx; dv = (2x − 5)−1/2 dx, v = (2x − 5)1/2 1 = x(2x − 5)1/2 − (2x − 5)1/2 dx = x(2x − 5)1/2 − (2x − 5)3/2 + C 3
4
4
CHAPTER 7. TECHNIQUES OF INTEGRATION
414 3.
1 u = ln 4x, du = dx; dv = dx, v = x x = x ln 4x − dx = x ln 4x − x + C
ln 4x dx
4.
ln(x + 1) dx
u = ln(x + 1), du =
1 dx; x+1
dv = dx, v = x
x dx = x ln(x + 1) − x+1
= x ln(x + 1) −
1−
1 x+1
dx
= x ln(x + 1) − x + ln(x + 1) + C 5.
u = ln 2x, du =
x ln 2x dx =
6.
7.
8.
x1/2 ln x dx
ln x dx x2
1 2 x ln 2x − 2
1 x2 1 dx = x2 ln 2x − 2x 2 2
1 2 x 2
x dx =
1 dx; dv = x1/2 dx, v = x 2 2 x3/2 2 = x3/2 ln x − dx = x3/2 ln x − 3 3 x 3 4 3/2 2 3/2 = x ln x − x + C 3 9
1 2 1 x ln 2x − x2 + C 2 4
2 3/2 x 3 2 x1/2 dx 3
1 1 1 u = ln x, du = dx; dv = 2 dx, v = − x x x 1 1 1 1 − 2 dx = − ln x − + C = − ln x − x x x x
ln x √ dx x3
u = ln x, du = −1/2
ln x −
= −2x−1/2 ln x + 2
9.
dv = x dx, v =
u = ln x, du =
= −2x
1 dx; x
(ln t)2 dt
1 dx; x
dv = x−3/2 dx, v = −2x−1/2
−2x−1/2 dx x x−3/2 dx = −2x−1/2 ln x − 4x−1/2 + C
u = (ln t)2 , du =
2 ln t dt; t
dv = dt, v = t
2 u = 2 ln t, du = dt; dv = dt, v = t t = t(ln t)2 − 2t ln t − 2 dt = t(ln t)2 − 2t ln t + 2t + C 2
= t(ln t) −
2 ln t dt
7.3. INTEGRATION BY PARTS 10.
(t ln t)2 dt
415
u = (ln t)2 , du =
2 ln t dt; t
dv = t2 dt, v =
1 3 t 3
2 2 2 1 2 t ln t dt u = ln t, du = dt; dv = t2 dt, v = t3 3 t 3 9 2 3 1 2 2 1 2 2 t ln t − t dt = t3 (ln t)2 − t3 ln t + t3 + C = t3 (ln t)2 − 3 9 9 3 9 27 =
11.
1 3 t (ln t)2 − 3
sin−1 x dx
u = sin−1 x, du = √
1 dx; 1 − x2
dv = dx, v = x
x dx u = 1 − x2 , du = −2x dx 1 − x2 √ 1 1 √ − du = x sin−1 x + u + C = x sin−1 x − 2 u −1 = x sin x + 1 − x2 + C = x sin−1 x −
12.
13.
14.
x2 tan−1 x dx
xe3x dx
√
1 1 dx; dv = x2 dx, v = x3 1 + x2 3 x3 x 1 3 1 1 1 3 −1 −1 x− = x tan x − dx = x tan x − dx 3 3 1 + x2 3 3 1 + x2 1 1 1 = x3 tan−1 x − x2 + ln(1 + x2 ) + C 3 6 6 u = tan−1 x, du =
1 u = x, du = dx; dv = e3x dx, v = e3x 3 1 1 3x 1 1 e dx = xe3x − e3x + C = xe3x − 3 3 3 9
x2 e5x dx
u = x2 , du = 2x dx;
dv = e5x dx, v =
1 5x e 5
2 2 1 x, du = dx; dv = e5x dx, v = e5x 5 5 5 2 5x 1 2 5x 1 2 2 5x xe − e dx = x2 e5x − xe5x + e +C = x2 e5x − 5 25 25 5 25 125 =
1 2 5x x e − 5
2 5x xe dx 5
u=
CHAPTER 7. TECHNIQUES OF INTEGRATION
416 15.
3 −4x
x e
1 3 3 3 −4x e dx = − x3 e−4x − x2 e−4x − xe−4x − +C 4 16 32 128
x3 3x2
e−4x
+
1 − e−4x 4
–
6x
+
6
–
1 −4x e 16 −
1 −4x e 64
1 −4x e 256
x5 5 x
5 x
4 x
3 x
2 x
x e dx = x e − 5x e + 20x e − 60x e + 120xe − 120e + C
16.
x
x
5x4
17.
18.
2
x3 ex dx
5 2x3
x e
2 1 2 u = x2 , du = 2x dx; dv = xex dx, v = ex 2 2 2 2 1 1 1 2 = x2 ex − xex dx = x2 ex − ex + C 2 2 2
1 3 dx u = x , du = 3x dx; dv = x e dx, v = e2x 6 1 3 2x3 1 2 2x3 1 3 2x3 1 2x3 x e dx = x e = x e − − e +C 6 2 6 12
19.
t cos 8t dt
3
2
2 2x3
20x3 60x2
+ – + –
120x
+
120
–
ex ex ex ex ex ex ex
1 u = t, du = dt; dv = cos 8t dt, v = sin 8t 8 1 1 1 1 sin 8t dt = t sin 8t + cos 8t + C = t sin 8t − 8 8 8 64
20.
21.
x sinh x dx
u = x, du = dx; dv = sinh x dx, v = cosh x = x cosh x − cosh x dx = x cosh x − sinh x + C
x2 sin x dx
u = x2 , du = 2x dx; dv = sin x dx, v = − cos x = −x2 cos x + 2x cos x dx u = 2x, du = 2 dx; dv = cos x dx, v = sin x = −x2 cos x + 2x sin x − 2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C
7.3. INTEGRATION BY PARTS 22.
23.
x2 cos
x dx 2
x3 cos 3x dx
x x u = x2 , du = 2x dx; dv = cos dx, v = 2 sin 2 2 x x = 2x2 sin − 4 x sin dx 2 2 x x u = x, du = dx; dv = sin dx, v = −2 cos 2 2 x x x = 2x2 sin − 4 −2x cos + 2 cos dx 2 2 2 x x x = 2x2 sin − 4 −2x cos + 4 sin +C 2 2 2 x x x = 2x2 sin + 8x cos − 16 sin + C 2 2 2 1 u = x3 , du = 3x2 dx; dv = cos 3x dx, v = sin 3x 3 1 = x3 sin 3x − x2 sin 3x dx 3 1 u = x2 , du = 2x dx; dv = sin 3x dx, v = − cos 3x 3 1 3 1 2 2 x cos 3x dx = x sin 3x − − x cos 3x + 3 3 3 1 dv = cos 3x dx, v = sin 3x 3 1 1 3 1 2 2 1 x sin 3x − sin 3x dx = x sin 3x + x cos 3x − 3 3 3 3 3 2 1 1 2 cos 3x + C = x3 sin 3x + x2 cos 3x − x sin 3x − 3 3 9 27 u = x, du = dx;
24.
x4 sin 2x dx
1 u = x4 , du = 4x3 dx; dv = sin 2x dx, v = − cos 2x 2 1 = − x4 cos 2x + 2 x3 cos 2x dx 2 1 u = x3 , du = 3x2 dx; dv = cos 2x dx, v = sin 2x 2 1 3 1 4 3 x sin 2x − x2 sin 2x dx = − x cos 2x + 2 2 2 2 1 dv = sin 2x dx, v = − cos 2x 2 1 4 1 2 3 = − x cos 2x + x sin 2x − 3 − x cos 2x + x cos 2x dx 2 2 u = x2 , du = 2x dx;
417
CHAPTER 7. TECHNIQUES OF INTEGRATION
418
1 dv = cos 2x dx, v = sin 2x 2 1 1 1 4 3 2 3 x sin 2x − sin 2x dx = − x cos 2x + x sin 2x + x cos 2x − 3 2 2 2 2 3 1 3 3 = − x4 cos 2x + x3 sin 2x + x2 cos 2x − x sin 2x − cos 2x + C 2 2 2 4 u = x, du = dx;
ex sin 4x dx
25.
u = sin 4x, du = 4 cos 4x dx; = ex sin 4x − 4ex cos 4x dx
dv = ex dx, v = ex
u = cos 4x, du = −4 sin 4x dx; dv = 4ex dx, v = 4ex = ex sin 4x − 4ex cos 4x + 16 ex sin 4x dx Solving for the integral, we have 17
ex sin 4x dx = ex sin 4x − 4ex cos 4x + C ex sin 4x dx =
26.
e−x cos 5x dx
or
ex (sin 4x − 4 cos 4x) + C. 17
1 u = e−x , du = −e−x dx; dv = cos 5x dx, v = sin 5x 5 1 1 1 1 e−x sin 5x dx = e−x sin 5x − − e−x sin 5x dx = e−x sin 5x + 5 5 5 5 1 u = e−x , du = −e−x dx; dv = sin 5x dx, v = − cos 5x 5 1 −x 1 1 1 −x −x e cos 5x dx = e sin 5x + − e cos 5x − 5 5 5 5 1 1 1 = e−x sin 5x − e−x cos 5x − e−x cos 5x dx 5 25 25
Solving for the integral, we have
26 25
1 −x 1 e sin 5x − e−x cos 5x + C 5 25 5 sin 5x − cos 5x e−x cos 5x dx = + C. 26ex e−x cos 5x dx =
or
7.3. INTEGRATION BY PARTS 27.
e−2θ cos θ dθ
419
u = e−2θ , du = −2e−2θ dθ; = e−2θ sin θ − −2e−2θ sin θ dθ
dv = cos θ dθ, v = sin θ
u = e−2θ , du = −2e−2θ dθ; dv = sin θ dθ, v = − cos θ = e−2θ sin θ + 2 −e−2θ cos θ − 2e−2θ cos θ dθ = e−2θ sin θ − 2e−2θ cos θ − 4 e−2θ cos θ dθ sin θ − 2 cos θ + C. Solving for the integral, we have e−2θ cos θ dθ = 5e2θ eαx sin βx dx
28.
u = eαx , du = αeαx dx; 1 = − eαx cos βx − β
dv = sin βx dx, v = −
1 cos βx β
α − eαx cos βx dx β
1 sin βx β 1 α 1 αx α αx e sin βx − e sin βx dx = − eαx cos βx + β β β β 2 α 1 α = 2 eαx sin βx − eαx cos βx − 2 eαx sin βx dx β β β eαx (α sin βx − β cos βx) Solving for the integral, we have eαx sin βx dx = + C. α2 + β 2 29. θ sec θ tan θ dθ u = θ, du = dθ; dv = sec θ tan θ dθ, v = sec θ = θ sec θ − sec θ dθ = θ sec θ − ln | sec θ + tan θ| + C u = eαx , du = αeαx dx;
30.
e2t cos et dt
dv = cos βx dx, v =
u = et , du = et dt; dv = et cos et dt, v = sin et = et sin et − et sin et dt = et sin et + cos et + C
31.
sin x cos 2x dx
u = cos 2x, du = −2 sin 2x dx; dv = sin x dx, v = − cos x = − cos x cos 2x − 2 cos x sin 2x dx u = 2 sin 2x, du = 4 cos 2x dx; dv = cos x dx, v = sin x = − cos x cos 2x − 2 sin x sin 2x − 4 sin x cos 2x dx
CHAPTER 7. TECHNIQUES OF INTEGRATION
420 Solving for the integral, we have
sin x cos 2x dx =
1 2 cos x cos 2x + sin x sin 2x + C. 3 3
32.
cosh x cosh 2x dx
u = cosh 2x, du = 2 sinh 2x dx; dv = cosh x dx, v = sinh x = sinh x cosh 2x − 2 sinh x sinh 2x dx u = 2 sinh 2x, du = 4 cosh 2x dx; dv = sinh x dx, v = cosh x = sinh x cosh 2x − 2 sinh 2x cosh x − 4 cosh x cosh 2x dx
Solving for the integral, we have 33.
34.
x3
x2 + 4 dx
t5 dx (t3 + 1)2
35.
sin(ln x) dx
cosh x cosh 2x dx =
1 2 sinh 2x cosh x − sinh x cosh 2x + C. 3 3
1 u = x2 , du = 2x dx; dv = x x2 + 4 dx, v = (x2 + 4)3/2 3 1 2 x(x2 + 4)3/2 dx = x2 (x2 + 4)3/2 − 3 3 2 1 = x2 (x2 + 4)3/2 − (x2 + 4)5/2 + C 3 15
1 u = t3 , du = 3t2 dt; dv = t2 (t3 + 1)−2 dt, v = − (t3 + 1)−1 3 1 3 3 1 1 = − t (t + 1)−1 + t2 (t3 + 1)−1 dt = − t3 (t3 + 1)−1 + ln |t3 + 1| + C 3 3 3
1 u = sin(ln x), du = cos(ln x) dx; x = x sin(ln x) − cos(ln x) dx
dv = dx, v = x
1 u = cos(ln x), du = − sin(ln x) dx; dv = dx, v = x x = x sin(ln x) − x cos(ln x) + sin(ln x) dx Solving for the integral, we have 36.
cos ln(sin x) dx
sin(ln x) dx =
1 1 x sin(ln x) − x cos(ln x) + C. 2 2
cos x dx; dv = cos x dx, v = sin x u = ln(sin x), du = sin x = sin x ln(sin x) − cos x dx = sin x ln(sin x) − sin x + C
7.3. INTEGRATION BY PARTS
csc3 x dx
37.
421
u = csc x, du = − csc x cot x dx; dv = csc2 x dx, v = − cot x = − csc x cot x − csc x cot2 x dx = − csc x cot x − csc x(csc2 x − 1) dx 3 = − csc x cot x − csc x dx + csc x dx = − csc x cot x − csc3 x dx + ln | csc x − cot x|
Solving for the integral, we have
x sec−1 x dx
38.
= = = =
x sec2 x dx
39.
1 1 u = sec−1 x, du = √ dx; dv = x dx, v = x2 2 2 x x −1 x2 1 2 1 1 1 2 −1 −1 √ x sec x − x(x2 − 1)−1/2 dx dx = x sec x − 2 2 x x2 − 1 2 2 1 2 1 x sec−1 x − (x2 − 1)−1/2 (2x dx) t = x2 − 1, dt = 2x dx 2 4 1 2 1 1 1 x sec−1 x − t−1/2 dt = x2 sec−1 x − (2t1/2 ) + C 2 4 2 4 1 2 1 2 1 2 −1 −1 x sec x − x − 1 + C = (x sec x − x2 − 1) + C 2 2 2
u = x, du = dx; dv = sec2 x dx, v = tan x = x tan x − tan x dx = x tan x − ln | sec x| + C
x tan2 x dx =
40.
1 1 csc3 x dx = − csc x cot x + ln | csc x − cot x| + C. 2 2
x(sec2 x − 1) dx =
x sec2 x dx −
x dx
u = x, du = dx; dv = sec2 x dx, v = tan x 1 1 = x tan x − tan x dx − x2 = x tan x − ln | sec x| − x2 + C 2 2
2
41.
x ln(x + 1) dx 0
1 1 dx; dv = x dx, v = x2 x+1 2 2 2 2 x 1 1 = x2 ln(x + 1) − dx 2 2 x +1 0 0 1 2 1 1 x−1+ dx = (4 ln 3 − 0) − 2 2 0 x+1 2
3 1 1 1 2 x − x + ln(x + 1)
= 2 ln 3 − (ln 3 − 0) = ln 3 = 2 ln 3 − 2 2 2 2 0 u = ln(x + 1), du =
CHAPTER 7. TECHNIQUES OF INTEGRATION
422
1
42.
ln(x2 + 1) dx
0
u = ln(x2 + 1), du =
dv = dx, v = x
1 2x2 1 dx = (ln 2 − 0) − 2 1 − dx 2 x2 + 1 0 x +1 0 π 1
π − (0 − 0) = + ln 2 − 2 = ln 2 − 2 x − tan−1 x 0 = ln 2 − 2 1 − 4 2
= x ln(x2 + 1)
2x dx; x2 + 1
4
43.
xe−x/2 dx
1
− 0
dv = e−x/2 dx, v = −2e−x/2
u = x, du = dx;
2
= −2xe−x/2
4 2
−
4 2
1
(−2e−x/2 ) dx = −2(4e−2 − 2e−1 ) − 4e−x/2
4 2
8 8 12 8e − 12 4 = − 2 − 4(e−2 − e−1 ) = − 2 = e e e e e2
π
ex cos x dx
44. −π
u = cos x, du = − sin x dx; π π = ex cos x]−π + ex sin x dx
dv = ex dx, v = ex
−π
dv = ex dx, v = ex π π = [−eπ − (−e−π )] + ex sin x]−π − ex cos x dx −π π −π π x = e − e + (0 − 0) − e cos x dx u = sin x, du = cos x dx;
−π
π
Solving for the integral, we have
ex cos x dx =
−π
1
45.
tan−1 x dx
0
√
46. 0
2/2
1 −π (e − eπ ). 2
1 dx; dv = dx, v = x 1 + x2 1 1 1 π 1 x 2 − 0 − ln(1 + x = x tan−1 x 0 − dx = ) 2 4 2 0 1+x 0 π 1 π 1 = − (ln 2 − 0) = − ln 2 4 2 4 2
cos−1 x dx
u = tan−1 x, du =
u = cos−1 x, du = − √ √2/2
√
1 dx; 1 − x2
dv = dx, v = x
2/2 x = x cos x 0 − −√ dx 1 − x2 0 √ √2/2 1 2π −0 − (1 − x2 )−1/2 (−2x dx) = 8 2 0 √ t = 1 − x2 , dt = −2x dx; t−1/2 dt = 2t1/2 = 2 t = 2 1 − x2 −1
7.3. INTEGRATION BY PARTS
423
√
1 2π − 2 1 − x2 8 2
=
= 0
3
47. A =
(1 + ln x) dx e−1
u = 1 + ln x, du =
3
√
√2/2
1 dx; x
2π − 8
1 √ − 1 2
√ 2π 2 − +1 8 2
√ =
dv = dx, v = x 2
3
3
= (x + x ln x)]e−1 − dx = (3 + 3 ln 3) − (e−1 − e−1 ) − x]e−1 e−1 1 1 = 3 + 3 ln 3 − 3 − = 3 ln 3 + e e 0 1 48. A = − tan−1 x dx + tan−1 x dx −1
0
u = tan−1 x, du =
3
1/e
1 dx; 1 + x2
-1 1
dv = dx, v = x
1
1 1 x x −1 = − x tan x −1 − dx + x tan x − dx 2 2 0 −1 1 + x 0 1+x 0 1 1 π π 1 2 2 − 0 − ln(1 + x ) + =− 0− − ln(1 + x ) 4 2 4 2 −1 0 −1
0
0
π 1 π π 1 + (0 − ln 2) + − (ln 2 − 0) = − ln 2 4 2 4 2 2 5 2 ln x dx; 49. V = π (ln x)2 dx u = (ln x)2 , du = x 1 5 5 = πx(ln x)2 1 − π 2 ln x dx =
2
dv = dx, v = x
1
5
1
2 dx; dv = dx, v = x x 5 5 2 = π[5(ln 5) − 0] − π 2x ln x]1 − 2 dx 1 5 = 5π(ln 5)2 − π (10 ln 5 − 0) − 2x]1 = 5π(ln 5)2 − 10π ln 5 + 8π u = 2 ln x, du =
ln 3
50. V = 2π 0
x(3 − e ) dx = 6π
ln 3
x
0 x
x dx − 2π
ln 3
xex dx
3
0
u = x, du = dx; dv = e dx, v = ex ln 3 2 ln 3 x ln 3 x = 3πx 0 − 2π xe ]0 − e dx
2
1
0
ln 3 = 3π(ln 3)2 − 6π ln 3 + 4π = 3π(ln 3)2 − 2π (3 ln 3 − 0) − ex ]0
ln 3
CHAPTER 7. TECHNIQUES OF INTEGRATION
424
π
u = x, du = dx; dv = sin x dx, v = − cos x 0 π π π = 2π −x cos x]0 + cos x dx = 2π [−(−π − 0) + sin x]0 ] = 2π 2
51. V = 2π
x sin x dx
1
0
sin x 1 (− sin x) = − = − tan x cos x cos x π/4 π/4 2 2 s= 1 + tan x dx = 1 + (sec x − 1) dx =
52. y =
0
0
π/4
√
sec2
0
π/4
x dx = 0
| sec x| dx
On [0, π/4], sec x > 0, so we have
π/4
s= 0
π/4
sec x dx = ln |sec x + tan x| ]0
√ √ = ln | 2 + 1| − ln |1 + 0| = ln(1 + 2).
2 x x 2 1 tan−1 dx u = tan−1 , du = dx; dv = dx, v = x 2−0 0 2 2 4 + x2 2 2 2 2x π 1 1 −1 x 2 = − dx = x tan 2 − ln(4 + x ) 2 2 2 0 2 4 0 0 4+x π 1 1 π − (ln 8 − ln 4) = − ln 2 = 2 2 4 2 54. s(t) = e−t sin t dt u = e−t , du = −e−t dt; dv = sin t dt, v = − cos t −t = −e cos t − e−t cos t dt u = e−t , du = −e−t dt; dv = cos t dt, v = sin t = −e−t cos t − e−t sin t + e−t sin t dt 53. fave =
Solving for the integral, we have s(t) =
1 1 e−t sin t dt = − e−t cos t − e−t sin t + C. Now 2 2
1 1 1 0 = s(0) = − + C, so C = and s(t) = (1 − e−t cos t − e−t sin t). 2 2 2 55. v(t) = te−t dt u = t, du = dt; dv = e−t dt, v = −e−t = −te−t + e−t dt = −te−t − e−t + C. Now 1 = v(0) = −1 + C, so C = 2 and v(t) = −te−t − e−t + 2. s(t) = (−te−t − e−t + 2) dt = − te−t dt + e−t + 2t = −(−te−t − e−t ) + e−t + 2t + C = te−t + 2e−t + 2t + C. Now −1 = s(0) = 2 + C, so C = −3 and s(t) = te−t + 2e−t + 2t − 3.
7.3. INTEGRATION BY PARTS
1
56. W = 62.4π 1/2 1
= 31.2π 1/2
425
x sin2 πx dx = 62.4π
1 1/2
x dx − 31.2π
x (1 − cos 2πx) dx 2
1
1 y
1/2
x cos 2πx dx
y
1
1/2
1 sin 2πx dv = cos 2πx dx, v = 2π 1 1 1 1 2 1 x sin 2πx sin 2πx = 15.6πx 1/2 − 31.2π − 2π 1/2 2π 1/2 1 1 1 = 15.6π 1 − cos 2πx − 31.2π 0 + 4 4π 2 1/2
x
u = x, du = dx;
15.6 7.8 (1 + 1) = 11.7π − ≈ 31.7910 ft-lb π π 57. Using symmetry, 2 4 πx πx πx dx dx, v = sin x cos u = x, du = dx; dv = cos F = 2 62.4 4 4 π 4 0 2 2 2 πx 8 16 4x 4 πx πx sin dx = 124.8 + 2 cos = 124.8 − sin π 4 0 π 0 4 π π 4 0 8 16 − 2 ≈ 115.4825 lb = 124.8 π π π/2 π/2 2 58. A = sin x dx = − cos x]0 = −(0 − 1) = 1 = 11.7π −
0
My =
1
π/2
x sin x dx 0
u = x, du = dx;
π/2
= −x cos x]0
π/2
+ 0
dv = sin x dx, v = − cos x π/2
cos x dx = −(0 − 0) + sin x]0
-1
=1
π/2 1 π/2 2 1 π/2 1 1 π π 1 sin x dx = (1 − cos 2x) dx = = Mx = x − sin 2x = 2 0 4 0 4 2 4 2 8 0 π 1 π/8 = x = = 1, y = 1 1 8 √ 4 −1 √ tan x 1 √ dx t = x, dt = √ dx 59. x 2 x 1 2 1 =2 tan−1 t dt u = tan−1 t, du = dt; dv = dt, v = t 1 + t2 1 2 2 t π 1 −1 2 −1 2 − ln(1 + t ) = 2 t tan t 1 − dt = 2 2 tan 2 − 2 4 2 1 1+t 1 = 4 tan−1 2 −
5 π π − (ln 5 − ln 2) = 4 tan−1 2 − − ln 2 2 2
CHAPTER 7. TECHNIQUES OF INTEGRATION
426 60.
xe
√
x
dx
√
x, x = t2 , dx = 2t dt = t2 et (2t dt) = 2 t3 et dt u = t3 , du = 3t2 dt; dv = et dt, v = et 3 t 2 t = 2 t e − 3t e dt u = t2 , du = 2t dt; dv = et dt, v = et 3 t 2 t t = 2t e − 6 t e − 2te dt u = t, du = dt; dv = et dt, v = et 3 t 2 t t t = 2t e − 6t e + 12 te − e dt = 2t3 et − 6t2 et + 12tet − 12et + C t=
= 2x3/2 e 61.
sin
√
√
x + 2 dx
x
− 6xe
t=
√
√
x
√ √ √ + 12 xe x − 12e x + C
x + 2, x = t2 − 2, dx = 2t dt
(sin t)2t dt u = 2t, du = 2 dt; dv = sin t dt, v = − cos t = −2t cos t + 2 cos t dt = −2t cos t + 2 sin t + C √ √ √ = −2 x + 2 cos x + 2 + 2 sin x + 2 + C =
π2
62.
cos
√
t dt
0
√
t, t = x2 , dt = 2x dx π π = cos x(2x dx) = 2 x cos x dx x=
0
0
u = x, du = dx; dv = cos x dx, v = sin x π π π = 2 x sin x]0 − sin x dx = 2 cos x]0 = 2(−1 − 1) = −4 0
(ln x)n dx
63.
n(ln x)n−1 dx; u = (ln x)n , du = x = x(ln x)n − n (ln x)n−1 dx
dv = dx, v = x
64.
sinn x dx
u = sinn−1 x, du = (n − 1) sinn−2 x cos x dx; dv = sin x dx, v = − cos x = − sinn−1 x cos x + (n − 1) cos2 x sinn−2 x dx = − sinn−1 x cos x + (n − 1) (1 − sin2 x) sinn−2 x dx = − sinn−1 x cos x + (n − 1) sinn−2 x dx − (n − 1) sinn x dx
7.3. INTEGRATION BY PARTS
427
Solving for the integral
sinn x dx, we have
sinn x dx = −
sinn−1 x cos x n − 1 + n n
sinn−2 x dx.
65.
u = cosn−1 x, du = −(n − 1) cosn−2 x sin x dx; dv = cos x dx, v = sin x = cosn−1 x sin x + (n − 1) sin2 x cosn−2 x dx n−1 = cos x sin x + (n − 1) (1 − cos2 x) cosn−2 x dx = cosn−1 x sin x + (n − 1) cosn−2 x dx − (n − 1) cosn x dx Solving for the integral cosn x dx, we have cosn x dx
cosn x dx =
cosn−1 x sin x n − 1 + n n
cosn−2 x dx.
66.
u = secn−2 x, du = (n − 2) secn−2 x tan x dx; dv = sec2 x dx, v = tan x = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx n−2 = sec x tan x − (n − 2) secn−2 x(sec2 x − 1) dx = secn−2 x tan x − (n − 2) secn x dx + (n − 2) secn−2 x dx Solving for secn x dx, we have secn x dx
secn x dx =
secn−2 x dx,
n = 1.
sinn−1 x cos x n − 1 + sinn−2 x dx with n = 3, n n sin2 x cos x 2 sin2 x cos x 2 3 + sin x dx = − − cos x + C. sin x dx = − 3 3 3 3
sinn x dx = −
67. Using
68. Using
secn−2 x tan x n − 2 + n−1 n−1
secn−2 x tan x n − 2 + secn−2 x dx with n = 4, n−1 n−1 sec2 x tan x 2 sec2 x tan x 2 4 + sec2 x dx = + tan x + C. sec x dx = 3 3 3 3
secn x dx =
CHAPTER 7. TECHNIQUES OF INTEGRATION
428 cosn x dx =
69. Using
cosn−2 x dx with n = 3,
u = 10x, du = 10 dx 2 1 cos2 u sin u cos3 u du = + cos u du = 10 10 · 3 10 · 3 cos2 u sin u 1 cos2 10x sin 10x 1 = + sin u + C = + sin 10x + C. 30 15 30 15
cosn−1 x sin x n − 1 + cos x dx = n n
n
70. Using
cos3 10x dx
cosn−1 x sin x n − 1 + n n
cosn−2 x dx with n = 4,
cos3 x sin x 3 cos3 x sin x 3 cos x sin x 1 2 0 + cos x dx = + + cos x dx cos x dx = 4 4 4 4 2 2 cos3 x sin x 3 cos3 x sin x 3 + + (cos x sin x + x) + C. = cos x sin x + dx = 4 8 4 8 4
sinn x dx = −
71. Using
π/2 0
sinn−1 x cos x n − 1 + n n
sinn−2 x dx, we have
π/2 sinn−1 x cos x n − 1 π/2 n−2 sin x dx = − + sin x dx n n 0 0 n−1 sin (π/2) cos(π/2) sinn−1 0 cos 0 n − 1 π/2 n−2 − sin x dx =− + n n n 0 n − 1 π/2 n−2 n − 1 π/2 n−2 sin x dx = sin x dx. = −(0 − 0) + n n 0 0 n
π/2
72. Repeated use of
sinn x dx =
0
π/2 0
n−1 n
π/2
sinn−2 x dx yields
0
k+3 k+1 n−1 n−3 · ··· · sin x dx = n n−2 k+4 k+2
π/2
n
sink x dx,
0
where k = 0 when n is even and n ≥ 2, and k = 1 when n is odd and n ≥ 3. Thus, we get, respectively:
π/2
(a)
sinn x dx =
0
3 1 n−1 n−3 · ··· · n n−2 4 2
n−1 n n−1 = n
=
n−3 3 ··· n−2 4 n−3 3 · ··· n−2 4 ·
π/2 0
sin0 x dx
1 π/2 dx 2 0 π 1 · 3 · 5 · · · (n − 1) 1 π · −0 = · . 2 2 2 2 · 4 · 6···n ·
7.3. INTEGRATION BY PARTS
π/2
(b)
n−1 n n−1 = n n−1 = n
sinn x dx =
0
π/2
73.
π/2
74. 75.
76.
sin8 x dx =
0
0
(sin−1 x)2 dx
n−3 4 ··· n−2 5 n−3 4 · ··· n−2 5 n−3 4 · ··· n−2 5 ·
2 π/2 1 sin x dx 3 0 2 π/2 · (− cos x)]0 3 2 2 · 4 · 6 · · · (n − 1) · [0 − (−1)] = . 3 3 · 5 · 7···n ·
105π 35π π 1·3·5·7 · = = 2 2·4·6·8 768 256
sin5 x dx = ·
e2x tan−1 ex dx
429
8 2·4 = 3·5 15 ex 1 dx; dv = e2x dx, v = e2x 1 + e2x 2 3x e 1 1 = e2x tan−1 ex − t = ex , dt = ex dx dx 2 2 1 + e2x t2 1 1 1 1 2x 1 −1 x e 1 − = e2x tan−1 ex − dt = tan e − dt 2 2 1 + t2 2 2 1 + t2 1 1 1 1 1 = e2x tan−1 ex − t + tan−1 t + C = (e2x + 1) tan−1 ex − ex + C 2 2 2 2 2 u = tan−1 ex , du =
2 sin−1 x u = (sin−1 x)2 , du = √ dx; 1 − x2 2x sin−1 x √ = x(sin−1 x)2 − dx 1 − x2
dv = dx, v = x
2x dx, v = −2 1 − x2 1 − x2 = x(sin−1 x)2 − −2 1 − x2 sin−1 x − −2 dx = x(sin−1 x)2 + 2 1 − x2 sin−1 x − 2x + C u = sin−1 x, du = √
77.
xex dx (x + 1)2
u = xex , du = (x + 1)ex dx; x x e + =− x+1
78.
x2 e x dx (x + 2)2
1 dx; 1 − x2
x
e dx =
dv =
x 1− x+1
u = x2 ex , du = x(x + 2)ex dx;
dv = √
1 1 dx, v = − (x + 1)2 x+1
ex + C = dv =
1 x ex e +C = +C x+1 x+1
1 1 dx, v = − (x + 2)2 x+2
1 2 x x e + xex dx =− u = x, du = dx; dv = ex dx, v = ex x+2 x2 x x2 x−2 x x x e + xe − e dx = x − 1 − e +C =− ex + C = x+2 x+2 x+2
CHAPTER 7. TECHNIQUES OF INTEGRATION
430 ex sin x dx:
79. We first compute
ex sin x dx
u = ex , du = ex dx; dv = sin x dx, v = − cos x = −ex cos x + ex cos x dx u = ex , du = ex dx; dv = cos x dx, v = sin x = −ex cos x + ex sin x − ex sin x dx. ex sin x dx =
Solving for the integral, we have
1 x e (sin x − cos x). Similarly, 2
ex cos x dx =
1 x e (sin x + cos x). Then 2 xex sin x dx
1 dv = ex sin x dx, v = ex (sin x − cos x) 2 1 x 1 (ex sin x − ex cos x) dx = xe (sin x − cos x) − 2 2 1 x 1 1 x 1 x = xe (sin x − cos x) − e (sin x − cos x) − e (sin x + cos x) + C 2 2 2 2 1 1 1 = xex sin x − xex cos x + ex cos x + C. 2 2 2 u = x, du = dx;
80. We first compute
e−x cos 2x dx:
e−x cos 2x dx
1 u = e−x , du = −e−x dx; dv = cos 2x dx, v = sin 2x 2 1 −x 1 e−x sin 2x dx = e sin 2x + 2 2 1 u = e−x , du = −e−x dx; dv = sin 2x dx, v = − cos 2x 2 1 −x 1 1 1 −x −x e cos 2x dx . = e sin 2x + − e cos 2x − 2 2 2 2
Solving for the integral, we have
e−x cos 2x dx =
1 −x e (2 sin 2x − cos 2x). Similarly, 5
7.3. INTEGRATION BY PARTS
431
1 e−x sin 2x dx = − e−x (sin 2x + 2 cos 2x). Then 5 1 u = x, du = dx; dv = e−x cos 2x dx, v = ex (2 sin 2x − cos 2x) xe−x cos 2x dx 5 1 1 (2e−x sin 2x − e−x cos 2x) dx = xe−x (2 sin 2x − cos 2x) − 5 5 1 2 1 = xe−x (2 sin 2x − cos 2x) − − e−x (sin 2x + 2 cos 2x) 5 5 5 1 1 −x + e (2 sin 2x + cos 2x) + C 5 5 1 4 3 2 −x = xe sin 2x − xe−x cos 2x + e−x sin 2x + e−x cos 2x + C. 5 5 25 25 81. ln x + x2 + 1 dx u = ln x + = x ln x +
x2
x2
+1
+1 −
= x ln x + x2 + 1 −
√ 1 + x/ x2 + 1 √ , du = dx; dv = dx, v = x x + x2 + 1 1 x √ √ 1+ x dx x + x2 + 1 x2 + 1 √ √ x2 + 1 + x x − x2 + 1 √ x dx 2 2 x − (x + 1) x2 + 1 x 2 2 √ x +1−x x +1+x dx x2 + 1 x √ dx 2 x +1
= x ln x + x2 + 1 − 2 = x ln x + x + 1 − = x ln x + x2 + 1 − x2 + 1 + C 82.
esin
−1
x
dx
u = esin = xe
sin−1 x
−1
−1
x
esin x , du = √ dx; 1 − x2
dv = dx, v = x
−1
xesin x √ dx 1 − x2
−
esin x x , du = √ dx; dv = √ dx, v = − 1 − x2 2 2 1−x 1−x sin−1 x sin−1 x sin−1 x 2 = xe − −e 1 − x − −e dx −1 −1 −1 = xesin x + esin x 1 − x2 − esin x dx u = esin
−1
−1
x
Solving for the integral, we have
esin
−1
−1
x
dx =
esin 2
x
(x +
1 − x2 ) + C.
CHAPTER 7. TECHNIQUES OF INTEGRATION
432 83. (a) Graph shown at right.
3
(b) We use the reduction formula sinn−1 x cos x n − 1 sinn x dx = − + sinn−2 x dx n n with n = 4 and n = 2. 2π A= (3 + sin2 x − 5 sin4 x) dx 0 =
2π 3x]0
2π
+2 0
2
sin x dx − 5
sin3 x cos x − 4
2π 0
0 0 7 7 2π 2 = 6π + 5 − + sin x dx = 6π − − 4 4 4 0 4 2π 17π 7 . = = 6π − x 8 0 4
3 + 4
2π
2
sin x dx 0
sin x cos x − 2
2π 0
π/4
π/4
dv = (sin x − cos x) dx, v = − cos x − sin x 5π/4 5π/4 = −x(cos x + sin x)]π/4 − −(cos x + sin x) dx u = x, du = dx;
5π 4 √ 3π 2 = 2 = −
π/4
√ √ √ π 2 2 2 2 5π/4 − − + + + (sin x − cos x)]π/4 2 2 4 2 2 √ √ √ √ √ 2 2 2 2 3π 2 + − + − − = 2 2 2 2 2
PROBLEMAS 7.4 Powers2.4 of Trigonometric Functions 1.
sin1/2 x cos x dx
u = sin x, du = cos x dx =
2.
cos4 5x sin 5x dx =
u1/2 du =
2π
dx 0
y = x sin x
(b) For x > 0, the first and second points of intersection of y = x sin x and y = x cos x are x1 = π/4 and x2 = 5π/4. 5π/4 5π/4 A= (x sin x − x cos x) dx = x(sin x − cos x) dx
√
1 + 2
3
84. (a) Graph shown at right.
2
2 3/2 2 u + C = (sin x)3/2 + C 3 3
u = cos 5x, du = −5 sin 5x dx 1 1 1 4 u − du = − u5 + C = − cos5 5x + C 5 25 25
-3
y = x cos x
7.4. POWERS OF TRIGONOMETRIC FUNCTIONS 3.
3
cos x dx =
cos x cos x dx =
= sin x − 4.
5.
6.
5
3
1 2 cos3 t − cos5 t + C 3 5
5
cos x sin x cos x dx = (1 − sin2 x) sin3 x cos x dx 1 1 = sin3 x(cos x dx) − sin5 x(cos x dx) = sin4 x − sin6 x + C 4 6
3
2
3
2
4
2
sin 2x cos 2x sin 2x dx =
9.
2
1 2 sin3 t + sin5 t + C 3 5
=
2
cos t cos t dt = (1 − sin t) cos t dt = (1 − 2 sin2 t + sin4 t) cos t dt = cos t dt − 2 sin2 t(cos t dt) + sin4 t(cos t dt) 4
sin 2x cos 2x dx =
8.
sin2 x(cos x dx)
1 sin3 x + C 3
sin x cos x dx =
cos x dx −
= sin t −
7.
(1 − sin x) cos x dx =
sin4 t sin t dt = (1 − cos2 t)2 sin t dt = (1 − 2 cos2 t + cos4 t) sin t dt = sin t dt + 2 cos2 t(− sin t dt) − cos4 t(− sin t dt)
sin5 t dt =
cos t dt =
2
sin2 4x sin 4x dx = (1 − cos2 4x) sin 4x dx 1 1 1 cos2 4x(−4 sin 4x dx) = cos3 4x − cos 4x + C = sin 4x dx + 4 12 4
sin3 4x dx =
= − cos t +
2
433
(1 − cos2 2x)2 cos2 2x sin 2x dx
(1 − 2 cos2 2x + cos4 2x) cos2 2x sin 2x dx
2 1 − cos 2t 1 sin t dt = (sin t) dt = (1 − 2 cos 2t + cos2 2t) dt dt = 2 4 3 1 + cos 4t 1 1 1 1 − 2 cos 2t + − 2 cos 2t + cos 4t dt = dt = 4 2 4 2 2 1 1 3 sin 4t + C = t − sin 2t + 8 4 32 4
2
2
CHAPTER 7. TECHNIQUES OF INTEGRATION
434 10.
11.
12.
13.
14.
3 1 + cos 2θ (cos2 θ)3 dθ = dθ 2 1 = (1 + 3 cos 2θ + 3 cos2 2θ + cos3 2θ) dθ 8 1 + cos 4θ 1 2 = 1 + 3 cos 2θ + 3 + (1 − sin 2θ) cos 2θ dθ 8 2 5 3 1 2 + 4 cos 2θ + cos 4θ − sin 2θ cos 2θ dθ = 8 2 2 1 3 1 5 θ + sin 2θ + sin 4θ − sin3 2θ + C = 16 4 64 6 2 4 2 4 sin x cos x dx = (1 − cos x) cos x dx = (cos4 x − cos6 x) dx 2 3 1 + cos 2x 1 + cos 2x − dx = 2 2 1 [2(1 + 2 cos 2x + cos2 2x) − (1 + 3 cos 2x + 3 cos2 2x + cos3 2x)] dx = 8 1 (1 + cos 2x + cos2 2x − cos3 2x) dx = 8 1 1 + cos 4x 2 1 + cos 2x − = − (1 − sin 2x) cos 2x dx 8 2 1 1 1 2 − cos 4x + sin 2x cos 2x dx = 8 2 2 1 1 1 x− sin 4x + sin3 2x + C = 16 64 48 cos3 x cos2 x 1 − sin2 x −2 (cos x dx) − cos x dx dx = cos x dx = cos x dx = (sin x) sin2 x sin2 x sin2 x 1 − sin x + C = − csc x − sin x + C =− sin x 2 1 − cos 4x 1 1 4 4 4 sin x cos x dx = sin 2x dx = dx 16 16 2 1 1 1 + cos 8x 2 = (1 − 2 cos 4x + cos 4x) dx = 1 − 2 cos 4x + dx 64 64 2 3 1 1 − 2 cos 4x + cos 8x dx = 64 2 2 1 1 3 x− sin 4x + sin 8x + C = 128 128 1024 2 1 1 − cos 12x 1 1 sin2 3x cos2 3x dx = sin 6x dx = sin2 6x dx = dx 2 4 4 2 1 1 1 1 = dx − cos 12x dx = x − sin 12x + C 8 8 8 96 cos6 θ dθ =
7.4. POWERS OF TRIGONOMETRIC FUNCTIONS
4
3
tan 2t sec 2t dt =
2
2
tan 2t sec 2t sec 2t dt = tan3 2t(1 + tan2 2t) sec2 2t dt 1 1 = tan3 2t(2 sec2 2t dt) + tan5 2t(2 sec2 2t dt) 2 2 1 1 tan6 2t + C = tan4 2t + 8 12 √ 2 2 2 1/2 2 16. (2 − tan x) sec x dx = 4 sec x dx − 4(tan x) sec x dx + tan x sec2 x dx
15.
3
435
1 8 = 4 tan x − (tan x)3/2 + tan2 x + C 3 2 17. tan2 x sec3 x dx = (sec2 x − 1) sec3 x dx = sec5 x dx − sec3 x dx u = sec3 x, du = 3 sec2 x sec x tan x dx; dv = sec2 x dx, v = tan x = tan x sec3 x − 3 tan2 x sec3 x dx − sec3 x dx 1 1 From Example 5 of Section 2.3, 7.3, sec3 x dx = sec x tan x + ln | sec x + tan x| + C, so 2 2 1 1 tan2 x sec3 x dx = tan x sec3 x − 3 tan2 x sec3 x dx − sec x tan x − ln | sec x + tan x|. 2 2 Solving for the integral, we have 1 1 1 tan2 x sec3 x dx = tan x sec3 x − sec x tan x − ln | sec x + tan x| + C. 4 8 8 1 tan2 3x sec2 3x(3 dx) 18. tan2 3x sec2 3x dx = u = tan 3x, du = sec2 3x(3 dx) 3 1 1 1 u2 du = u3 + C = tan3 3x + C = 3 9 9 19. tan3 x(sec x)−1/2 dx = tan2 x(sec x)−1/2 tan x dx = (sec2 x − 1)(sec x)−3/2 sec x tan x dx 1/2 = (sec x) (sec x tan x dx) − (sec x)−3/2 (sec x tan x dx) 2 (sec x)3/2 + 2(sec x)−1/2 + C 3 x x x x x x x x x x sec2 − 1 sec2 sec tan dx 20. tan3 sec3 dx = tan2 sec2 sec tan dx = 2 2 2 2 2 2 2 2 2 2 x x x x x x = sec4 sec tan dx − sec2 sec tan dx 2 2 2 2 2 2 2 2 5 x 3 x = sec − sec +C 5 2 3 2 =
CHAPTER 7. TECHNIQUES OF INTEGRATION
436 21.
3
5
= 22.
tan x sec x sec x tan x dx = (sec2 x − 1) sec4 x sec x tan x dx = sec6 x(sec x tan x dx) − sec4 x(sec x tan x dx)
tan x sec x dx =
5
tan x sec x dx =
2
4
1 1 sec7 x − sec5 x + C 7 5 2
2
(tan x) (sec x tan x dx) =
(sec2 x − 1)2 (sec x tan x dx)
(sec4 x − 2 sec2 x + 1)(sec x tan x dx) 4 2 = sec x(sec x tan x dx) − 2 sec x(sec x tan x dx) + sec x tan x dx
=
2 1 sec5 x − sec3 x + sec x + C 5 3 23. sec5 x dx = sec3 x sec2 x dx = sec3 x(1 + tan2 x) dx = sec3 x dx + tan2 x sec3 x dx 1 1 2.3, From Example 5 of Section 7.3, sec3 x dx = sec x tan x + ln | sec x + tan x| + C. From 2 2 1 1 1 2 3 3 Exercise 17, tan x sec x dx = tan x sec x − sec x tan x − ln | sec x + tan x| + C1 . 4 8 8 1 1 sec x tan x + ln | sec x + tan x| Thus, sec5 x = 2 2 1 1 1 + tan x sec3 x − sec x tan x − ln | sec x + tan x| + C2 4 8 8 3 1 3 = sec x tan x + ln | sec x + tan x| + tan x sec3 x + C2 . 8 8 4 1 dx = sec2 x sec2 x dx = (1 + tan2 x) sec2 x dx 24. cos4 x 1 = sec2 x dx + tan2 x sec2 x dx = tan x + tan3 x + C 3 =
(1 − sin2 x) cos x dx sin x 1 = (sin x)−1 (cos x dx) − sin x(cos x dx) = ln | sin x| − sin2 x + C 2 1 or (Alternatively, sin x cos x dx = cos x(sin x dx) = − cos2 x + C 2 1 1 sin 2x dx = − cos 2x + C). sin x cos x dx = 2 4 1 1 26. sin x sec7 x dx = (cos x)−7 (sin x dx) = (cos x)−6 + C = sec6 x + C 6 6
25.
cos2 x cot x dx =
cos3 x dx = sin x
7.4. POWERS OF TRIGONOMETRIC FUNCTIONS 27.
cot
10
4
10
cot x csc x csc x dx = cot10 x(1 + cot2 x) csc2 x dx = − cot10 x(− csc2 x) dx − cot12 x(− csc2 x) dx
x csc x dx =
2
437
2
1 1 cot13 x + C = − cot11 x − 11 13 2 2 2 2 2 28. (1 + csc t) dt = (1 + 2 csc t + csc t csc t) dt = [1 + 2 csc2 t + (1 + cot2 t) csc2 t] dt 1 = (1 + 3 csc2 t + cot2 t csc2 t) dt = t − 3 cot t − cot3 t + C 3 2 4 2 sec (1 − t) sec (1 − t) 1 + tan (1 − t) dt = sec2 (1 − t) dt = sec2 (1 − t) dt 29. 8 8 tan (1 − t) tan (1 − t) tan8 (1 − t) −8 2 = [tan(1 − t)] sec (1 − t) dt + [tan(1 − t)]−6 sec2 (1 − t) dt 1 1 [tan(1 − t)]−7 + [tan(1 − t)]−5 + C 7 5 1 1 + +C = 7 tan7 (1 − t) 5 tan5 (1 − t) =
30.
sin3
√ √ t cos2 t √ dt t
u= =2
√
1 t, du = √ 2 t
sin3 u cos2 u du = 2
sin2 u cos2 u sin u du
(1 − cos2 u) cos2 u sin u du = 2 cos2 u(sin u du) − 2 cos4 u(sin u du) =2
√ √ 2 2 2 2 = − cos3 u + cos5 u + C = − cos3 t + cos5 t + C 3 5 3 5 2 31. (1 + tan x) sec x dx = (1 + 2 tan x + tan2 x) sec x dx = sec x dx + 2 tan x sec x dx + tan2 x sec x dx The last integral is evaluated in Example 8 of Section 2.4, 7.4. Thus, 1 1 (1 + tan x)2 sec x dx = ln | sec x + tan x| + 2 sec x + sec x tan x − ln | sec x + tan x| + C 2 2 1 1 = ln | sec x + tan x| + 2 sec x + sec x tan x + C. 2 2 32. (tan x + cot x)2 dx = (tan2 x + 2 + cot2 x) dx = (sec2 x − 1 + 2 + csc2 x − 1) dx = (sec2 x + csc2 x) dx = tan x − cot x + C
CHAPTER 7. TECHNIQUES OF INTEGRATION
438 33.
4
tan x dx =
tan5 x dx =
2
tan2 x(sec2 x − 1) dx
tan x tan x dx =
=
34.
2
tan2 x sec2 x dx −
(tan2 x)2 tan x dx =
(sec2 x − 1) dx =
1 tan3 x − tan x + x + C 3
(sec2 x − 1)2 tan x dx
(sec4 x − 2 sec2 x + 1) tan x dx = sec3 x(sec x tan x dx) − 2 sec x(sec x tan x dx) + tan x dx
=
1 sec4 x − sec2 x − ln | cos x| + C 4 35. cot3 t dt = cot2 t cot t dt = (csc2 t − 1) cot t dt = csc t(csc t cot t dt) − cot t dt =
1 = − csc2 t − ln | sin t| + C 2 36.
csc5 t dt
u = csc3 t, du = −3 csc2 t cot t csc t dt; dv = csc2 t dt, v = − cot t 3 2 3 3 = − cot t csc t − 3 cot t csc t dt = − cot t csc t − 3 (csc2 t − 1) csc3 t dt 2.3, Problem 37 = − cot t csc3 t − 3 csc5 t dt + 3 csc3 t dt See Exercises 7.3, 1 1 3 5 = − cot t csc t − 3 csc t dt + 3 − cot t csc t + ln | csc t − cot t| 2 2 3 3 = − cot t csc3 t − cot t csc t ln | csc t − cot t| − 3 csc5 t dt 2 2
Solving for the integral, we have
37.
6
3 1 3 csc5 t dt = − cot t csc3 t − cot t csc t + ln | csc t − cot t| + C. 4 8 8
2
(tan x − tan x) dx =
=
4
2
2
(tan x tan x − tan x) dx =
(tan4 x − 1) tan2 x dx
(tan4 x − 1)(sec2 x − 1) dx
(tan4 x sec2 x − tan4 x − sec2 x + 1) dx = tan4 x(sec2 x dx) − tan4 x dx − sec2 x dx + dx 1 = tan5 x − tan4 x dx − tan x + x 5
=
7.4. POWERS OF TRIGONOMETRIC FUNCTIONS From Exercise 33,
38.
tan4 x =
439
1 tan3 x − tan x + x + C, so 3 1 1 tan5 x − tan3 x − tan x + x − tan x + x + C1 5 3 1 1 = tan5 x − tan3 x + C1 . 5 3
(tan6 x − tan2 x) dx =
cot 2x csc5/2 2x dx =
csc3/2 2x csc 2x cot 2x dx = −
1 2
(csc3/2 2x)(−2 csc 2x cot 2x dx)
u = csc 2x, du = −2 csc 2x cot 2x dx 1 1 2 5/2 1 3/2 u du = − u =− + C = − csc5/2 2x + C 2 2 5 5
3
2
2
2
x sin x sin x dx = x(1 − cos2 x2 ) sin x2 dx = x sin x2 dx − x cos2 x2 sin x2 dx 1 1 sin x2 (2x dx) + cos2 x2 (−2x sin x2 dx) = 2 2
x sin x dx =
39.
2
t = x2 , dt = 2x dx; u = cos x2 , du = −2x sin x2 dx 1 1 1 = sin t dt + u2 du = − cos t + u3 + C 2 2 3 1 1 1 1 = − cos x2 + cos3 x2 + C = cos3 x2 − cos x2 + C 2 3 6 2 40.
1 tan8 x2 (2x sec2 x2 dx) u = tan x2 , du = 2x sec2 x2 dx 2 1 1 1 9 1 8 u du = u +C = tan9 x2 + C = 2 2 9 18
x tan8 x2 sec2 x2 dx =
π/2
41. π/3
√ sin3 θ cos θ dθ =
π/2
sin2 θ(cos θ)1/2 sin θ dθ =
π/3
=−
π/2
(cos θ) π/3
1/2
π/2
(1 − cos2 θ)(cos θ)1/2 sin θ dθ
π/3 π/2
(− sin θ dθ) + π/3
(cos θ)5/2 (− sin θ dθ)
π/2 π/2 2 2 3/2 7/2 = − (cos θ) + (cos θ) 3 7 π/3 π/3 √ √ 3/2 7/2 √ 1 1 25 2 2 2 2 2 0− − = + = =− 0− 3 2 7 2 6 56 168
CHAPTER 7. TECHNIQUES OF INTEGRATION
440
π/2
42.
sin5 x cos5 x dx =
0
π/2
0
π/2 0
π/2
=
sin5 x cos4 x cos x dx =
0 π/2
= 0
sin5 x(1 − sin2 x)2 cos x dx
sin5 x(1 − 2 sin2 x + sin4 x) cos x dx sin5 x(cos x dx) − 2
π/2
sin7 x(cos x dx) +
0 π/2
π/2
sin9 x(cos x dx)
0
π/2 1 1 1 1 1 1 8 10 sin x = − sin x + = − + 4 10 6 4 10 60 0 0 0 π π π sin3 2t dt = sin2 2t sin 2t dt = (1 − cos2 2t) sin 2t dt 43. 0 0 0 π π π 1 π 1 1 = sin 2t dt + cos2 2t(−2 sin 2t dt) = − cos 2t + cos3 2t 2 0 2 6 0 0 0 1 1 = − (1 − 1) + (1 − 1) = 0 2 6 π π π 44. sin4 x cos2 x dx = sin4 x(1 − sin2 x) dx = (sin4 x − sin6 x) dx −π −π −π 2 3 π 1 − cos 2x 1 − cos 2x dx − = 2 2 −π 1 π [2(1 − 2 cos 2x + cos2 2x) − (1 − 3 cos 2x + 3 cos2 2x − cos3 2x)] dx = 8 −π 1 π = (1 − cos 2x − cos2 2x + cos3 2x) dx 8 −π 1 π 1 + cos 4x 2 = + (1 − sin 2x) cos 2x dx 1 − cos 2x − 8 −π 2 π 1 1 1 2 = − cos 4x − sin 2x cos 2x dx 8 −π 2 2 π 1 1 1 1 π 1 3 = x − sin 4x − sin 2x (π + π) = = 8 2 8 6 16 8 −π π/4 π/4 π/4 45. tan y sec4 y dy = tan y sec2 y sec2 y dy = tan y(1 + tan2 y) sec2 y dy 1 = sin6 x 6
0
π/2
0 π/4
= 0
tan y sec2 y dy +
0
π/4 0
tan3 y sec2 y dy
π/4 π/4 1 3 1 1 1 tan2 y + tan4 y = (1 − 0) + (1 − 0) = 2 4 2 4 4 0 0 π/3 π/3 π/3 2 3/2 1/2 3/2 tan x sec x dx = sec x(sec x tan x dx) = sec x 46. 3 0 0 0 √ √ 2 4 2−2 = (2 2 − 1) = 3 3 =
7.4. POWERS OF TRIGONOMETRIC FUNCTIONS
47.
48. 49. 50.
51.
1 [sin 3x + sin(−x)] dx = (sin 3x − sin x) dx 2 1 1 = − cos 3x + cos x + C 6 2 1 1 1 (cos 2x + cos 8x) dx = sin 2x + sin 8x + C cos 3x cos 5x dx = 2 4 16 1 1 1 sin 2x sin 4x dx = (cos 2x − cos 6x) dx = sin 2x − sin 6x + C 2 4 12 5 − 3 sin 2x 5 3 dx = (5 cos 6x − 3 sin 2x cos 6x) dx = sin 6x − [sin 8x + sin(−4x)] dx sec 6x 6 2 5 3 5 3 3 = sin 6x − (sin 8x − sin 4x) dx = sin 6x + cos 8x − cos 4x + C 6 2 6 16 8 π/6 π/6 1 π/6 1 1 cos 2x cos x dx = (cos x + cos 3x) dx = sin x + sin 3x 2 0 2 3 0 0 1 1 1 5 + = = 2 2 3 12 1 sin x cos 2x dx = 2
π/2
52. 0
441
π/2 1 1 π/2 3 1 1 sin x sin x dx = (cos x − cos 2x) dx = sin x − sin 2x 2 2 2 0 2 2 0 1 1 = (1 − 0) = 2 2
53. If m = n, then using the fact that sin mx sin nx is an even function we have π π sin mx sin nx dx = [cos(m − n)x − cos(m + n)x] dx −π 0 π π 1 1 = sin(m − n)x − sin(m + n)x = 0. m−n m+n 0 0 If m = n, then π π π π sin mx sin nx dx = sin2 mx dx = 2 sin2 mx dx = (1 + cos 2mx) dx −π −π 0 0 π 1 = x+ sin 2mx = π. 2m 0
π
sin mx sin nx dx =
Thus, −π
0, π,
m = n . m=n
π
54. Since sin mx cos nx is an odd function,
sin mx sin nx dx = 0. −π
CHAPTER 7. TECHNIQUES OF INTEGRATION
442
55. y = sec2
x 2
2
π/2 x x x V =π dx = π sec sec2 sec2 dx 2 2 2 −π/2 −π/2 π/2 x x =π 1 + tan2 sec2 dx 2 2 −π/2 π/2 π/2 1 1 x x x sec2 tan2 = 2π dx + sec2 dx 2 2 2 2 2 −π/2 −π/2 π/2 x π/2 x 1 1 1 = 2π tan = 2π 1 − (−1) + − (−1) + tan3 2 −π/2 3 2 −π/2 3 3 2 16π = 2π 2 + = 3 3 π/2
y
4
–/2
1
56. y = sin2 x
1–y
π/2
V =π
/2
−π/2
(1 − sin2 x)2 dx = π
2
π/2
–/2
(cos2 x)2 dx
/2
−π/2
1 + cos 2x π π/2 dx = (1 + 2 cos 2x + cos2 2x) dx 2 4 −π/2 −π/2 1 π π/2 1 + cos 4x π π/2 3 + 2 cos 2x + cos 4x dx = 1 + 2 cos 2x + dx = 4 −π/2 2 4 −π/2 2 2 π/2 π 3 1 3π 3π 3π 2 π = x + sin 2x + sin 4x +0+0 − − +0+0 = = 4 2 8 4 4 4 8 −π/2 π/2
=π
1
π/4
57. A =
−3π/4 π/4
=
−3π/4 π/4
= −3π/4
(cos3 x − sin3 x) dx (cos2 x cos x − sin2 x sin x) dx =
-
-1
π/4 −3π/4
[(1 − sin2 x) cos x − (1 − cos2 x) sin x] dx
[cos x − (sin2 x) cos x − sin x + (cos2 x) sin x] dx
π/4 1 1 3 3 = sin x − sin x + cos x − cos x 3 3 −3π/4 √ √ √ √ √ √ √ √ √ √ √ 5 2 2 2 2 2 2 2 2 2 2 − + − − − + − + =2 2− = = 2 12 2 12 2 12 2 12 3 3
7.5. TRIGONOMETRIC SUBSTITUTIONS 58. A = =
1 2 1 8
2π 0
0
2π
r2 dθ =
1 2
2π
sin2 4θ sin2
0
443
1 θ dθ = 2 8
2π 0
(1 − cos 8θ)(1 − cos θ) dθ
(1 − cos θ − cos 8θ + cos 8θ cos θ) dθ
2π 1 1 (cos 7θ + cos 9θ) dθ + θ − sin θ − sin 8θ 8 16 0 2π 1 1 1 1 π = (2π) + sin 7θ + sin 9θ = 8 16 7 9 4 0 π π 3 59. Based on the graphs, the values of cos x dx, cos5 x dx, 0 0 π cos7 x dx all appear to be 0. We note that, for every t and 0 π π π − t = − cos + t , thus lending such that 0 ≤ t ≤ , cos 2 2 2 credence to this conjecture. 1 = 8
0
π
π
n
cos x dx = 0
1
-1 -2
60. Based on Problem 59, we conjecture that the value of π cosn x dx: integer, is 0. To prove this, we evaluate
2
π
cos3 x cos5 x cos7 x
cosn x dx, where n is a positive odd
0
cosn−1 x cos x dx
0
Since n is a positive odd integer, n − 1 is guaranteed to be even, and thus k = integer. Further, we substitute cos2 x = 1 − sin2 x: π π π cosn x dx = (cos2 x)k cos x dx = (1 − sin2 x)k cos x dx. 0
0
n−1 is an 2
0
2
By the binomial theorem, (1 − sin x) expands into an expression of the form 1 + c1 sin2 x + c2 (sin2 x)2 + c3 (sin2 x)3 + · · · + ck (sin2 x)k . For this particular proof, it is not necessary to specify the precise values of the binary coefficients ci . Using the u substitution u = sin x, du = cos x dx, integration is accomplished as follows: π π n cos x dx = (1 + c1 sin2 x + c2 sin4 x + · · · + ck sin2k x)(cos x dx) 0 0 π c1 c2 ck 3 5 2k+1 sin x + sin x + · · · + sin = sin x + x 3 5 2k + 1 0 π cosn x dx = 0. Evaluating this, we note that sin π = sin 0 = 0, and therefore k
0
7.5
Trigonometric Substitutions
PROBLEMAS 2.5
CHAPTER 7. TECHNIQUES OF INTEGRATION
444
√ 1.
√
2.
1 − x2 dx x2
1
x = sin θ, dx = cos θ dθ cos2 θ 1 − sin2 θ cos θ dθ = dθ = cot2 dθ = sin2 θ sin2 θ √ 1 − x2 2 − sin−1 x + C = (csc θ − 1) dθ = − cot θ − θ + C = − x
x = 2 sec θ, dx = 2 sec θ tan θ dθ
8 sec θ (2 sec θ tan θ) dθ = 8 sec4 θ dθ 4 sec2 θ − 4 1 = 8 (1 + tan2 θ) sec2 θ dθ = 8 tan θ + tan3 θ + C 3 1 = 4 x2 − 4 + (x2 − 4)3/2 + C 3
√
1 − x2
x 1 dx = cosh−1 + C = ln x + x2 − 362 + C, 6 x2 − 36
√
x2 − 4
θ
3
√
=
3.
θ
√
x
x3 dx x2 − 4
x
2
x>6
Alternatively, the substitution x = 6 sec θ could have been used.
4.
5.
x
6.
3−
x2
dx
(1 − x2 )3/2 dx
1 2
3 sin θ, dx =
√
√
3
3 cos θ dθ θ √ 3 − x2 2 2 = 3 − 3 sin θ 3 cos θ dθ = 3 cos θ dθ 1 + cos 2θ 3 3 3 1 =3 dθ = θ + sin 2θ + C = θ + sin θ cos θ + C 2 2 2 2 2 √ 2 x 3−x x 3 x 3 3 x √ √ = sin−1 √ + + C = sin−1 √ + 3 − x2 + C 2 2 3 2 3 3 3 2
x2 + 7 dx =
x=
√
(x2 + 7)1/2 (2x dx) =
x
1 2 (x + 7)3/2 + C 3
1
x = sin θ, dx = cos θ dθ θ
√
1 − x2
x
7.5. TRIGONOMETRIC SUBSTITUTIONS
445
2 1 + cos 2θ (1 − sin2 θ)3/2 cos θ dθ = cos4 θ dθ = dθ 2 1 1 1 + cos 4θ = (1 + 2 cos 2θ + cos2 2θ) dθ = 1 + 2 cos 2θ + dθ 4 4 2 3 1 3 1 1 1 + 2 cos 2θ + cos 4θ dθ = θ + sin 2θ + sin 4θ + C = 4 2 2 8 4 32 1 1 3 1 3 sin 2θ cos 2θ + C = θ + sin 2θ(4 + cos 2θ) + C = θ + sin 2θ + 8 4 16 8 16 3 1 = θ + sin θ cos θ(4 + 1 − 2 sin2 θ) + C 8 8 1 3 = θ + sin θ cos θ(5 − 2 sin2 θ) + C 8 8 3 1 −1 = sin x + x 1 − x2 (5 − 2x2 ) + C 8 8 1 x2 1 − x2 (−2x dx) u = 1 − x2 , x2 = 1 − u, 2x dx = −du 7. x3 1 − x2 dx = − 2 1 1 1/2 (1 − u)u du = − (u1/2 − u3/2 ) du =− 2 2 1 2 3/2 2 5/2 1 1 =− u − u + C = − (1 − x2 )3/2 + (1 − x2 )5/2 + C 2 3 5 3 5 1 x2 x2 − 1(2x dx) u = x2 − 1, x2 = u + 1, 2x dx = du 8. x3 x2 − 1 dx = 2 1 1 1 2 5/2 2 3/2 1/2 3/2 1/2 (u + 1)u du = (u + u ) du = u + u = +C 2 2 2 5 3 1 1 = (x2 − 1)5/2 + (x2 − 1)3/2 + C 5 3 1 9. dx x = 2 sec θ, dx = 2 sec θ tan θ dθ x √ (x2 − 4)3/2 x2 − 4 2 sec θ tan θ θ = dθ (4 sec2 θ − 4)3/2 2 sec θ 1 1 −2 (sin θ) cos θ dθ = dθ = 4 4 tan2 θ 1 x 1 1 +C = − (sin θ)−1 + C = − csc θ + C = − √ 4 4 4 x2 − 4 10. (9 − x2 )3/2 dx x = 3 sin θ, dx = 3 cos θ dθ 3 x 2 3/2 = (9 − 9 sin θ) (3 cos θ dθ) θ 9 − x2 = −81 (1 − sin2 θ)3/2 (− cos θ dθ) = 81 cos4 θ dθ
=
= 81
2
2
(cos θ) dθ = 81
1 + cos 2θ 2
2 dθ
CHAPTER 7. TECHNIQUES OF INTEGRATION
446 = = = = = = = = = 11. x2 + 4 dx
1 + cos 4θ 81 81 2 (1 + 2 cos 2θ + cos 2θ) dθ = 1 + 2 cos 2θ + dθ 4 4 2 3 1 81 3 1 81 + 2 cos 2θ + cos 4θ dθ = θ + sin 2θ + sin 4θ + C 4 2 2 4 2 8 81 81 243 θ+ sin 2θ + sin 4θ + C 8 4 32 243 81 81 θ+ sin θ cos θ + sin 2θ cos 2θ + C 8 2 16 243 81 81 θ+ sin θ cos θ + sin θ cos θ(1 − 2 sin2 θ) + C 8 2 8 1 1 1 243 2 θ + 81 sin θ cos θ + − sin θ + C 8 2 8 4 5 1 243 2 θ + 81 sin θ cos θ − sin θ + C 8 8 4 5 1 2 243 −1 x 2 sin + 9x 9 − x − x +C 8 3 8 36 x 1 243 sin−1 + x 9 − x2 (45 − 2x2 ) + C 8 3 8 x = 2 tan θ, dx = 2 sec2 θ dθ
=
2
2
4 tan θ + 4 2 sec θ dθ = 4
4 + x2
3
sec θ dθ
x
θ
2
See Section 7.3, 2.3, Example 5 = 2 sec θ tan θ + 2 ln | sec θ + tan θ| + C
√ √
x2 + 4 x
x2 + 4 x
+ +C =2 + 2 ln
2 2 2 2
x 2
= x + 4 + 2 ln x2 + 4 + x + C1 2 1 x 1 1 12. dx = (2x dx) = ln(25 + x2 ) + C 25 + x2 2 25 + x2 2 x 1 √ dx = sin−1 + C 13. 5 25 − x2 |x| 1 1 √ +C 14. dx = sec−1 5 5 x x2 − 25 1 √ 15. dx x = 4 sin θ, dx = 4 cos θ dθ x 16 − x2 4 cos θ 1 csc θ dθ = dθ = 4 4 sin θ 16 − 16 sin2 θ
1 4 16 − x2
1 +C = ln | csc θ − cot θ| + C = ln
− 4 4 x x
4
√
θ
16 − x2
x
7.5. TRIGONOMETRIC SUBSTITUTIONS 16.
x2
√
1 dx 16 − x2
447
x = 4 sin θ, dx = 4 cos θ dθ
4
4 cos θ 1 csc2 θ dθ = = 2 2 16 16 sin θ 16 − 16 sin θ √ 1 16 − x2 1 +C = − cot θ + C = − 16 16 x 17.
18.
1 √ dx x 1 + x2
√
√ 20.
21.
x = tan θ, dx = sec2 θ dθ sec θ sec2 θ √ dθ = dθ = csc θ dθ = tan θ tan θ 1 + tan2 θ
√
1 + x2 1
= ln | csc θ − cot θ| + C = ln
− +C
x x
1 √ dx x2 1 + x2
19.
1 − x2 dx x4
x2 − 1 dx x4
√
√
x
θ
16 − x2
1 + x2
x
θ
1
√ x = tan θ, dx = sec2 θ dθ 1 + x2 sec2 θ sec θ θ √ = dθ dθ = tan2 θ 1 tan2 θ 1 + tan2 θ √ 2 1+x +C = (sin θ)−2 cos θ dθ = −(sin θ)−1 + C = − csc θ + C = − x
x = sin θ, dx = cos θ dθ cos2 θ 1 − sin2 θ cos θ dθ = dθ = cot2 θ csc2 θ dθ = sin4 θ sin4 θ √ 3 1 − x2 1 1 1 3 = − cot θ + C = − + C = − 3 (1 − x2 )3/2 + C 3 3 x 3x x = sec θ, dx = sec θ tan θ dθ tan2 θ tan θ sec θ tan θ dθ = dθ = sin2 θ cos θ dθ = sec4 θ sec3 θ √ 3 2−1 x 1 1 1 = sin3 θ + C = + C = 3 (x2 − 1)3/2 + C 3 3 x 3x
x2 dx (9 − x2 )3/2
x = 3 sin θ, dx = 3 cos θ dθ
2
2
sin θ dθ = tan2 θ dθ cos2 θ x x − sin−1 + C = (sec2 θ − 1) dθ = tan θ − θ + C = √ 3 9 − x2
=
9 sin θ 3 cos θ dθ = (9 − 9 sin2 θ)3/2
1
x
x
θ
√
1 − x2
x
√
x2 − 1
θ
1
3 θ 9 − x2
x
CHAPTER 7. TECHNIQUES OF INTEGRATION
448 22.
x2 dx (4 + x2 )3/2
4 + x2
x = 2 tan θ, dx = 2 sec2 θ dθ
2
x
2
θ 4 tan θ tan θ dθ 2 sec2 θ dθ = 2 2 3/2 sec θ (4 + 4 tan θ) 2 sec θ − 1 dθ = sec θ dθ − cos θ dθ = ln | sec θ + tan θ| − sin θ + C = sec θ
√
4 + x2
x
x x
+ − √ = ln
+ C = ln 4 + x2 + x − √ + C1 2
2 2
4+x 4 + x2
=
23.
1 dx (1 + x2 )2
24.
(x2
√
x = tan θ, dx = sec2 θ dθ 1 + x2 x 2 sec θ 1 θ dθ = cos2 θ dθ = dθ = 1 sec2 θ (1 + tan2 θ)2 1 1 1 1 1 (1 + cos 2θ) dθ = θ + sin 2θ + C = θ + sin θ cos θ + C = 2 2 4 2 2 1 1 x x 1 1 1 −1 −1 √ √ = tan x + + C = tan x + +C 2 2 2 2 1 + x2 1 + x2 1 + x2
x2 dx − 1)2
x = sec θ, dx = sec θ tan θ dθ =
2
sec θ sec θ tan θ dθ = (sec2 θ − 1)2
x
3
sec θ = tan3 θ
csc3 θ dθ
√
θ
1
u = csc θ, du = − csc θ cot θ dθ; dv = csc2 θ dθ, v = − cot θ 2 = − csc θ cot θ − csc θ cot θ dθ = − csc θ cot θ − csc θ(csc2 θ − 1) dθ = − csc θ cot θ − csc3 θ dθ + csc θ dθ = − csc θ cot θ − csc3 θ dθ + ln | csc θ − cot θ| Solving for the integral csc3 θ dθ, we have
1 1 csc3 θ dθ = − csc θ cot θ + ln | csc θ − cot θ| + C 2 2
x 1 1 x 1
1
+C √ √ =− + ln √ −√ 2 2 2 2 2 x −1 x −1 2 x −1 x − 1
1 x 1
x − 1
√ =− + ln √ + C. 2 2 x2 − 1 x2 − 1
x2 − 1
7.5. TRIGONOMETRIC SUBSTITUTIONS 25.
26.
1 dx (4 + x2 )5/2
x = 2 tan θ, dx = 2 sec2 θ dθ 1 1 2 sec2 θ 1 dθ = cos3 θ dθ = dθ = 16 sec3 θ 16 (4 + 4 tan2 θ)5/2 1 1 cos2 θ cos θ dθ = (1 − sin2 θ) cos θ dθ = 16 16 1 1 1 1 cos θ − sin2 θ cos θ dθ = sin θ − sin3 θ + C = 16 16 16 48 x x3 1 1 √ = − +C 16 48 (4 + x2 )3/2 4 + x2
x3 dx = (1 − x2 )5/2
4 + x2
x
θ
2
x2 (x dx) u = 1 − x2 , du = −2x dx (1 − x2 )5/2 1 1−u 1 (u−5/2 − u−3/2 ) du = − du = − 2 2 u5/2 1 1 2 −3/2 −1/2 + 2u =− + C = (1 − x2 )−3/2 − (1 − x2 )−1/2 + C − u 2 3 3
√
27.
449
1 dx = x2 + 2x + 10
1 (x + 1)2 + 9
√
dx
x2 + 2x + 10
x+1
θ
x + 1 = 3 tan θ, dx = 3 sec2 θ dθ 3 2 3 sec θ √ dθ = sec θ dθ = ln | sec θ + tan θ| + C = 2 9 tan θ + 9
√
x2 + 2x + 10 x + 1
= ln
+
+ C = ln x2 + 2x + 10 + x + 1 + C1
3 3
√
28.
x dx = 4x − x2
x dx = 4 − (x − 2)2
x−2 4 − (x − 2)2
dx +
2
dx
4 − (x − 2)2
u = 4 − (x − 2)2 , du = −2(x − 2) dx 1 x−2 x−2 1 = −u1/2 + 2 sin−1 +C =− du + 2 sin−1 1/2 2 2 2 u x−2 +C = − 4 − (x − 2)2 + 2 sin−1 2 29.
1 dx = (x2 + 6x + 13)2
1 dx [(x + 3)2 + 4]2
√
x2 + 6x + 13
x + 3 = 2 tan θ, dx = 2 sec2 θ dθ 1 2 sec2 θ sec2 θ dθ = cos2 θ dθ = dθ = 8 sec4 θ 8 (4 tan2 θ + 4)2
θ
2
x+3
CHAPTER 7. TECHNIQUES OF INTEGRATION
450
1 1 1 (1 + cos 2θ) dθ = = θ + sin 2θ + C 16 16 2 1 1 θ+ sin θ cos θ + C = 16 16 2 1 x+3 1 −1 x + 3 √ √ tan + = +C 2 2 16 2 16 x + 6x + 13 x + 6x + 13 x+3 x+3 1 tan−1 + +C = 16 2 8(x2 + 6x + 13) 1 1 30. dx = dx 6 (11 − 10x − x2 )2 [36 − (x + 5)2 ]2
31.
x−3 (5 − 4x − x2 )3/2
x+5
x + 5 = 6 sin θ, dx = 6 cos θ dθ θ √ 11 − 10x − x2 6 cos θ cos θ dθ = dθ = 216 cos4 θ (36 − 36 sin2 θ)2 1 sec3 θ dθ = See Section 7.3, 2.3, Example 5 216 1 1 1 sec θ tan θ + ln | sec θ + tan θ| + C = 216 2 2 x+5 6 1 √ √ = 2 432 11 − 10x − x 11 − 10x − x2
6 1 x+5
+C ln √ + +√ 2 432 11 − 10x − x2 11 − 10x − x
x+5 1 x + 11
+C = + ln √ 72(11 − 10x − x2 ) 432 11 − 10x − x2
x−3 dx = dx 3 [9 − (x + 2)2 ]3/2 x+2
x + 2 = 3 sin θ, dx = 3 cos θ dθ
√
θ 5 − 4x − x2
3 sin θ − 5 (3 cos θ dθ) (9 − 9 sin2 θ)3/2 9 sin θ cos θ − 15 cos θ dθ = 27 cos3 θ 1 sin θ 5 1 1 5 = dθ − dθ = tan θ sec θ dθ − sec2 θ dθ 3 cos2 θ 9 cos2 θ 3 9 5 1 = sec θ − tan θ + C 3 9 3 x+2 1 5 √ √ = − +C 3 9 5 − 4x − x2 5 − 4x − x2 −5x − 1 +C = √ 9 5 − 4x − x2 =
7.5. TRIGONOMETRIC SUBSTITUTIONS 32.
33. 34. 35. 36.
37.
38.
1 dx = (x2 + 2x)3/2
1 dx [(x + 1)2 − 1]3/2
451
x+1
x + 1 = sec θ, dx = sec θ tan θ dθ sec θ tan θ sec θ dθ = (sin θ)−2 cos θ dθ = dθ = (sec2 θ − 1)3/2 tan2 θ x+1 = −(sin θ)−1 + C = − csc θ + C = − √ +C x2 + 2x
√
x2 + 2x
θ
1
2x + 4 dx = ln(x2 + 4x + 13) + C x2 + 4x + 13 x−3 1 1 +C dx = tan−1 4 + (x − 3)2 2 2 x2 x 16 dx = 1 − dx = x − 4 tan−1 + C x2 + 16 x2 + 16 4 √ 2 4 − 9x dx 3x = 2 sin θ, 3 dx = 2 cos θ dθ 2 x 3x 2 4 − 4 sin θ 2 θ cos θ dθ = √ 2 4 − 9x2 3 sin θ 3 cos2 θ 1 − sin2 θ =2 dθ = 2 dθ sin θ sin θ = 2 csc θ dθ − 2 sin θ dθ = 2 ln | csc θ − cot θ| + 2 cos θ + C
√ √
2 4 − 9x2
4 − 9x2
− +C = 2 ln
+2
3x 3x 2
2 − √4 − 9x2
= 2 ln
+ 4 − 9x2 + C
3x 6x − x2 dx = 9 − (x − 3)2 dx x − 3 = 3 sin θ, dx = 3 cos θ dθ 3 x–3 = 9 − 9 sin2 θ 3 cos θ dθ = 9 cos2 θ dθ θ 6x − x2 9 9 1 = (1 + cos 2θ) dθ = θ + sin 2θ + C 2 2 2 √ 9 9 9 x−3 6x − x2 9 −1 x − 3 + +C = θ + sin θ cos θ + C = sin 2 2 2 3 2 3 3 x−3 1 9 + (x − 3) 6x − x2 + C = sin−1 2 3 2 1 1 √ u = x − 3, du = dx dx = dx 2 6x − x 9 − (x − 3)2 u x−3 1 √ +C = du = sin−1 + C = sin−1 2 2 3 3 3 −u
CHAPTER 7. TECHNIQUES OF INTEGRATION
452
1
39.
−1
4 − x2 dx
x = 2 sin θ, dx = 2 cos θ dθ
π/6
=
−π/6
2 4 − 4 sin θ 2 cos θ dθ = 4
π/6
π/6
(2 + 2 cos 2θ) dθ = (2θ + sin 2θ)
=
π/6 −π/6
−π/6
= √
3
40. −1
√ √ √ π 2π + 3 3 3 3 π + − − − = 3 2 3 2 3
x2 √ dx 4 − x2
π/3 −π/6
= 5
41. 0
4 sin2 θ 2 cos θ dθ −π/6 4 − 4 sin2 θ π/3 π/3 2 sin θ dθ = (2 − 2 cos 2θ) dθ = (2θ − sin 2θ)
x = 2 sin θ, dx = 2 cos θ dθ =4
1 dx (x2 + 25)3/2
−π/6
√ √ √ 3 3 π 2π − − − + =π− 3 3 2 3 2
2 √
2
x3
√
1 dx x2 − 1
−π/6
x = 5 tan θ, dx = 5 sec2 θ dθ
5 sec2 θ 1 dθ = 2 3/2 25 (25 tan θ + 25) 0 √ π/4 1 2 = sin θ = 25 50 0
42.
π/3
=
π/4
π/4
=
cos2 θ dθ
−π/6
0
1 1 dθ = sec θ 25
π/4
cos θ dθ 0
x = sec θ, dx = sec θ tan θ dθ π/3 π/3 sec θ tan θ tan θ √ dθ = cos2 θ dθ dθ = 2 θ tan θ 3 θ sec2 θ − 1 sec sec π/4 π/4 π/4 π/3 π/3 1 1 1 = (1 + cos 2θ) dθ = θ + sin2 θ 2 π/4 2 2 π/4 √ √ π 1 π 3 1 π+3 3−6 + − + = = 2 3 4 4 2 24
π/3
=
6/5
43. 1
x4
16 √ dx 4 − x2
x = 2 sin θ, dx = 2 cos θ dθ
sin−1 (3/5) 32 cos θ 2 cos θ dθ = dθ 4 4 2 sin (2 cos θ) π/6 π/6 16 sin θ 4 − 4 sin θ sin−1 (3/5) sin−1 (3/5) csc2 θ csc2 θ dθ = (1 + cot2 θ) csc2 θ dθ = sin−1 (3/5)
=
π/6
π/6
7.5. TRIGONOMETRIC SUBSTITUTIONS
sin−1 (3/5)
=
453
2
sin−1 (3/5)
csc θ dθ + π/6
cot2 θ csc2 θ dθ
π/6
sin−1 (3/5) 1 3 = − cot θ − cot θ 3 π/6 π/6 √ √ 1 = −[cot(sin−1 3/5) − 3] − [cot3 (sin−1 3/5) − 3 3] 3 √ √ 4 √ 1 64 172 − 3 − −3 3 =2 3− =− 3 3 27 81 sin−1 (3/5)
1/2
44.
x3 (1 + x2 )−1/2 dx
tan−1 (1/2)
=
tan−1 (1/2) 0 tan−1 (1/2)
= 0
=
tan3 θ(1 + tan2 θ)−1/2 sec2 θ dθ
0
=
tan3 θ sec2 θ dθ = sec θ
tan−1 (1/2)
tan2 θ tan θ sec θ dθ
0
(sec2 θ − 1) tan θ sec θ dθ
1 sec3 θ − sec θ 3
tan−1 (1/2) 0
√
1 = [sec3 (tan−1 1/2) − 1] − [sec(tan−1 1/2) − 1] 3 √ √ √ 16 − 7 5 5 1 5 5 −1 − −1 = = 3 8 2 24
x2 sin−1 x dx
u = sin−1 x, du = √ = = = = =
3
4
x = tan θ, dx = sec2 θ dθ
0
45.
5
1 dx; 1 − x2
dv = x2 dx, v =
5
2
1 3 x 3
1 θ
√ x3 1 − x2 dx x = sin θ, dx = cos θ dθ 2 1−x sin3 θ 1 3 −1 1 1 1 x sin x − sin3 θ dθ cos θ dθ = x3 sin−1 x − 3 3 3 3 1 − sin2 θ 1 3 −1 1 x sin x − (1 − cos2 θ) sin θ dθ 3 3 1 3 −1 1 1 3 x sin x − − cos θ + cos θ + C 3 3 3 1 1 1 3 −1 x sin x + 1 − x2 − (1 − x2 )3/2 + C 3 3 9
1 3 −1 1 x sin x − 3 3
√
1
x
CHAPTER 7. TECHNIQUES OF INTEGRATION
454 46.
x cos−1 x dx
u = cos−1 x, du = − √
1 dx; 1 − x2
dv = x dx, v =
1 2 x 2
1
x
θ
√ x2 1 − x2 √ dx x = sin θ, dx = cos θ dθ 2 1−x sin2 θ 1 1 1 1 sin2 θ dθ = x2 cos−1 x + cos θ dθ = x2 cos−1 x + 2 2 2 2 2 1 − sin θ 1 2 1 2 1 1 1 −1 −1 = x cos x + (1 − cos 2θ) dθ = x cos x + θ − sin 2θ + C 2 4 2 4 2 1 1 1 = x2 cos−1 x + sin−1 x − sin θ cos θ + C 2 4 4
1 1 = x2 cos−1 x + 2 2
√
2
√ √ 1 √ 47. A = dx x = 3 tan θ, dx = 3 sec2 θ dθ x 3 + x2 1 1 √ π/4 3 sec2 θ √ dθ = √ 3 tan θ 3 + 3 tan2 θ π/6 1 π/4 π/4 1 sec θ 1 =√ dθ = √ csc θ dθ 3 π/6 tan θ 3 π/6 √ π/4 √ √ 2−1 1 1 1 √ ≈ 0.2515 = √ ln | csc θ − cot θ| = √ (ln | 2 − 1| − ln |2 − 3|) = √ ln 3 3 3 2− 3 π/6
1
48. A =
3
x5
0
1 − x2 dx = 2
1 0
x4
1
1 − x2 x dx
2
u = 1 − x , x = 1 − u, 2x dx = −du 1 0 1/2 1 2 1/2 = (1 − u) u (u − 2u3/2 + u5/2 ) du − du = − 2 2 1 1 0 1 2 3/2 4 5/2 2 7/2 1 16 8 u − u + u ≈ 0.0762 =− =− 0− = 2 3 5 7 2 105 105 1
0
0
π/2
=4 0
= 2a
2
a2 − a2 sin2 θ a cos θ dθ = 4a2
π/2
(1 + cos 2θ) dθ = 2a 0
2
π/2 0
1 θ + sin 2θ 2
a
cos2 θ dθ
π/2
= 2a2
0
1
a
49. We find the area in the first quadrant and use symmetry. a A=4 a2 − x2 dx x = a sin θ, dx = a cos θ dθ
2
π 2
= πa2 a
50. We find the area in the first quadrant and use symmetry.
b
7.5. TRIGONOMETRIC SUBSTITUTIONS
455
a 2 b − x2 dx x = b sin θ, dx = b cos θ dθ 0 b π/2 4a π/2 2 2 2 = b − b sin θ b cos θ dθ = 4ab cos2 θ dθ b 0 0 π/2 π/2 π 1 + 0 = πab (1 + cos 2θ) dθ = 2ab θ + sin 2θ = 2ab = 2ab 2 2 0 0 b
A=4
√
√ √ 1 dx x = 3 tan θ, dx = 3 sec2 θ dθ 2 +x ) 0 √ √ π/4 π 3 π/4 1 3 sec2 θ =π dθ = 2 2 2 dθ 9 π/6 3 tan θ(3 + 3 tan θ) π/6 tan θ π/4 √ √ π/4 √ π 3 π/4 3 3 π π = (− cot θ − θ) cot2 θ dθ = (csc2 θ − 1) dθ = 9 9 9 π/6 π/6 π/6 √ √ π 3 π π 3 √ π π √ =− 3+ 3−1− 1+ − = ≈ 0.2843 9 4 6 9 12 3
51. V = π
2
x2 (3
1
√
3
1
2
1
52. Using the disk method, 2 16 dx x = 2 tan θ, dx = 2 sec2 θ dθ V =π (4 + x2 ) 2 0 π/4 π/4 π/4 16 sec2 θ 2 dθ = 2π =π (2 sec θ dθ) = 2π cos2 θ dθ sec4 θ (4 + 4 tan2 θ)2 0 0 0 π/4 π/4 π 1 1 π 2 + 2π =π + . (1 + cos 2θ) dθ = π θ + sin 2θ =π = 2 4 2 4 0 0 53. Using the shell method, 2 x2 4 + x2 dx V = 2π
0 π/4
= 2π
4 tan2 θ
π/4
= 32π 0
2
x = 2 tan θ, dx = 2 sec2 θ dθ
4 + 4 tan2 θ 2 sec2 θ dθ = 32π
0
1
(sec2 θ − 1) sec3 θ dθ = 32π
π/4
5
tan2 θ sec3 θ dθ
0
π/4 0
sec5 θ dθ − 32π
From Section 2.3, 7.3, Example 5 we obtain π/4 π/4 1 1 3 sec θ tan θ + ln | sec θ + tan θ| sec θ dθ = 2 2 0 0 √ √ 1 1 = ( 2)(1) + ln( 2 + 1) 2 2 √ 1√ = [ 2 + ln( 2 + 1)]. 2
π/4
sec3 θ dθ.
0
1
2
CHAPTER 7. TECHNIQUES OF INTEGRATION
456
π/4
To find
sec5 θ dθ we use integration by parts.
0
π/4
sec5 θ dθ =
0
π/4
sec3 θ sec2 θ dθ
0
u = sec3 θ, du = 3 sec2 θ sec θ tan θ dθ; π/4 π/4 3 = sec θ tan θ −3 sec3 θ tan2 θ dθ √ =2 2−3
0
√ =2 2−3 √ =2 2−3
π/4 0 π/4 0
dv = sec2 θ dθ, v = tan θ
0
sec3 θ(sec2 θ − 1) dθ sec5 θ dθ + 3
π/4
sec3 θ dθ
0
√ 3√ sec5 θ dθ + [ 2 + ln( 2 + 1)] 2 0 √ π/4 π/4 √ 2 3√ 5 5 + [ 2 + ln( 2 + 1)]. Then sec θ dθ we obtain sec θ dθ = Solving for 2 8 0 0 √ √ √ √ 2 3√ 2 1 + [ 2 + ln( 2 + 1)] − 32π + ln( 2 + 1) V = 32π 2 8 2 2 √ √ = 12π 2 − 4π ln( 2 + 1).
1
π/4
x2 dx x = 2 sin θ, dx = 2 cos θ dθ 4 − x2 0 π/6 π/6 4 sin2 θ = 2π sin2 θ dθ 2 cos θ dθ = 8π 0 0 4 − 4 sin2 θ π/6 π/6 1 (1 − cos 2θ) dθ = 4π θ − sin 2θ = 4π 2 0 0 √ √ 2 3 2π − 3π 3 π − −0 = ≈ 1.1383 = 4π 6 4 3
54. V = 2π
√
1
55. y = 1/x √3 √3 √ 2 x +1 L= dx 1 + (1/x)2 dx = x = tan θ, dx = sec2 θ dθ x 1 1 π/3 π/3 √ 2 π/3 tan θ + 1 sec3 θ sec θ 2 sec θ dθ = dθ = sec2 θ dθ = tan θ tan θ tan θ π/4 π/4 π/4 π/3 π/3 π/3 sin θ + csc θ dθ = csc θ(tan2 θ + 1) dθ = (sec θ tan θ + csc θ) dθ = cos2 θ π/4 π/4 π/4
π/3 √ √
2 1
= 2 + ln √ − √ − ( 2 + ln | 2 − 1|) = (sec θ + ln | csc θ − cot θ|) π/4 3 3
1
7.5. TRIGONOMETRIC SUBSTITUTIONS =2−
√
√ √ 2 − ln( 6 − 3) ≈ 0.9179
56. y = −x + 2 2 L= 1 + (2 − x)2 dx
1
0
= π/4
457
2 − x = tan θ, −dx = sec2 θ dθ
1 + tan2 θ(− sec2 θ dθ) =
0
sec3 θ dθ
2.3, Example 5 See Section 7.3,
π/4
π/4 √ 1 1√ 1 1 1 sec θ tan θ + ln | sec θ + tan θ| = 2(1) + ln | 2 + 1| − 0 − ln 1 2 2 2 2 2 0 √ √ 2 1 = + ln( 2 + 1) ≈ 1.1478 2 2 √ dy a2 − x2 , which is also . 57. (a) The slope at (x, y) is − x dx √ 2 a − x2 θ dx = − dy. Now (b) Separating variables, √ a x a 2 − x2 √ 2 2 a −x dx x = a sin θ, dx = a cos θ dθ x x √ 2 2 2 cos θ 1 − sin θ a − a2 sin θ a cos θ dθ = a dθ = a dθ = a sin θ sin θ sin θ = a csc θ dθ − a sin θ dθ = a ln | csc θ − cot θ| + a cos θ + C
√
a √a 2 − x 2
a2 − x2
+ C. = a ln −
+a
x x a
a − √a2 − x2 √
Then a ln
+ a2 − x2 = −y + C1 . Now y(10) = 0 and a = 10, so
x √
10 − 100 − 100 √
+ 100 − 100 = 0 + C1 and C1 = 0. Thus 10 ln
10
10 − √100 − x2
y = −10 ln
− 100 − x2 .
x
=
Note: If the substitution y = a cos θ is used, we obtain the equivalent solution
10 − √100 − x2
y = 10 ln
− 100 − x2 .
x 58. Using symmetry with respect to the x-axis, we have
CHAPTER 7. TECHNIQUES OF INTEGRATION
458
a+r
V = 4π
x
r2 − (x − a)2 dx r
a−r
x − a = r sin θ, dx = r cos θ dθ π/2 = 4π (a + r sin θ) r2 − r2 sin2 θ r cos θ dθ −π/2
a+r
π/2 1 + cos 2θ 3 dθ + 4πr (a + r sin θ) cos θ dθ = 4πar cos2 θ sin θ dθ = 4πr 2 −π/2 −π/2 −π/2 π/2 π/2 1 1 2 3 3 = 2πar θ + sin 2θ + 4πr − cos θ 2 3 −π/2 −π/2 π 4 π + 0 − − + 0 − πr3 (0 − 0) = 2aπ 2 r2 = 2πar2 2 2 3 2 2 2 2−x dx = 62.4 59. F = 62.4 x 2x − x2 dx = 62.4 1 − (x − 1)2 dx x 1 1 1 2
π/2
2
x − 1 = sin θ, dx = cos θ dθ π/2 = 62.4 1 − sin2 θ cos θ dθ = 62.4 0
π/2
π/2 0
cos2 θ dθ = 31.2
π/2 π 1 = 31.2 = 31.2 θ + sin 2θ = 15.6π ≈ 49.0088 lb 2 2 0 √3 1 √ A= dx x = tan θ, dx = sec2 θ dθ 2 1+x 0 π/3 π/3 sec2 θ √ dθ = = sec θ dθ 1 + tan2 θ 0 0 π/3 √ = ln | sec θ + tan θ| = ln(2 + 3)
60.
2
a–r
0
π/2
(1 + cos 2θ) dθ 0
1
1
√
3
√
3 √3 x 1 1 (1 + x2 )1/2 √ dx = 2x(1 + x2 )−1/2 dx = · My = 2 0 2 1/2 1 + x2 0 0 √3 = 1 + x2 =2−1=1
√
3
0
√3 1 1 π −1 dx = tan x = 2 1+x 2 6 0 0 π 1 π/6 √ ≈ 0.76; y = √ = √ ≈ 0.40 x= ln(2 + 3) ln(2 + 3) 6 ln(2 + 3) 1 √ 61. (a) dx ex = sec θ, ex dx = tan θ sec θ dθ, dx = tan θ dθ 2x e −1 tan θ tan θ √ dθ = dθ = θ + C = sec−1 ex + C = dθ = tan θ sec2 θ − 1 1 Mx = 2
√
3
2
7.5. TRIGONOMETRIC SUBSTITUTIONS (b)
e2x − 1 dx
459
ex = sec θ, ex dx = tan θ sec θ dθ, dx = tan θ dθ ex = sec2 θ − 1 tan θ dθ = tan2 θ dθ θ 1 = (sec2 θ − 1) dθ = tan θ − θ + C = e2x − 1 − sec−1 ex + C
e2x − 1
62. The circle of radius a, which is centered on the origin, is defined by x2 +y 2 = a2 . Let c be the distance between the centers of the two circles. The circle b c 2 2 2 of radius b is thus defined by x √ area is the integral √ + (y − c) = b . The a of the difference between y = b2 − x2 + c and y = a2 − x2 from −b to b. Using symmetry, we have b b [( b2 − x2 + c) − a2 − x2 ] dx = 2 ( b2 − x2 − a2 − x2 + c) dx A=2 0
0
By using the substitution x = r sin θ, dx = r cos θ dθ , we find 2 2 2 2 2 2 r − x dx = r − r sin θ r cos θ dθ = r cos2 θ dθ r2 r2 r2 1 + cos 2θ 1 dθ = sin θ cos θ + C = r2 θ + sin 2θ + C = θ + 2 2 2 2 2 √ r 2 x r 2 − x2 r2 x 1 r2 −1 x sin + = sin−1 + x r2 − x2 + C. = 2 r 2 r r 2 r 2 Substituting b and a respectively, we get b b b A=2 b2 − x2 dx − 2 a2 − x2 dx + 2 c dx
0
0
0
b b b x x + x b2 − x 2 = b sin − a2 sin−1 + x a2 − x2 + 2cx b a 0 0 0 b −1 −1 2 2 + b a2 − b2 − (0 + 0) + 2bc = b sin 1 + b b2 − b2 − (0 + 0) − a sin a 2 b πb − a2 sin−1 − b a2 − b2 + 2bc. = 2 a √ From the figure, it can be seen that a2 = b2 + c2 or c = a2 − b2 . We substitute to simplify further: πb2 b b πb2 − a2 sin−1 − b a2 − b2 + 2bc = − a2 sin−1 − bc + 2bc A= 2 a 2 a πb2 −1 b 2 = − a sin + b a 2 − b2 . 2 a 2
−1
The special-case lune of Hippocrates specifies a lune where the triangle formed by the origin √ 2 and the intersections of the two circles is a right isosceles triangle. For this lune, b = a 2 π b by the Pythagorean theorem and sin−1 is . Substituting these values above yields the a 4
CHAPTER 7. TECHNIQUES OF INTEGRATION
460
well-known result that the area of the lune of Hippocrates is the same as the area of the right 1 isosceles triangle that defines it, or a2 . 2
PROBLEMAS 2.6
7.6
Partial Fractions
1. Write
x−1 A B x−1 = = + . 2 x +x x(x + 1) x x+1
2. Write
A B 9x − 8 = + . (x − 3)(2x − 5) x − 3 2x − 5
3. Write
B C A D x3 + + = + . 3 2 (x − 1)(x + 2) x − 1 x + 2 (x + 2) (x + 2)3
4. Write
2x2 − 3 A B 2x2 − 3 C = + 2+ . = 3 2 2 x + 6x x (x + 6) x x x+6
5. Write
A C Dx + E 4 B = + 2+ 3+ 2 . x3 (x2 + 3) x x x x +3
6. Write
−x2 + 3x + 7 A B Cx + D = + . + 2 (x + 2)2 (x2 + x + 1) x + 2 (x + 2)2 x +x+1
7. Write
2x3 − x Cx + D Ax + B + 2 = 2 . 2 2 (x + 9) x +9 (x + 9)2
8.
3x2 − x + 4 3x2 − x + 4 = . x4 + 2x3 + x x(x3 + 2x2 + 1) This expression does not fall under any of the four partial fraction decomposition cases covered in Section 2.6, 7.6. 1 A B = + . Then 1 = A(x − 2) + Bx. x(x − 2) x x−2 Setting x = 0 and x = 2 gives A = −1/2 and B = 1/2. Thus 1 1 1 1 1 dx = − dx + dx x(x − 2) 2 x 2 x−2
1 1 1
x − 2
+ C. = − ln |x| + ln |x − 2| + C = ln
2 2 2 x
9. Write
1 A B = + . Then 1 = A(2x + 3) + Bx. x(2x + 3) x 2x + 3 Setting x = 0 and x = −3/2 gives A = 1/3 and B = −2/3. Thus 1 1 1 2 1 dx = dx − dx x(2x + 3) 3 x 3 2x + 3
1 1 1 x
+ C. = ln |x| − ln |2x + 3| + C = ln
3 3 3 2x + 3
10. Write
7.6. PARTIAL FRACTIONS
461
x+2 x+2 A B = = + . Then x + 2 = A(2x − 1) + Bx. 2x2 − x x(2x − 1) x 2x − 1 Setting x = 0 and x = 1/2 gives A = −2 and B = 5. Thus x+2 1 1 5 dx = −2 dx + 5 dx = −2 ln |x| + ln |2x − 1| + C. 2x2 − x x 2x − 1 2
11. Write
3x + 10 A B = + . Then 3x + 10 = A(x + 2) + Bx. 2 x + 2x x x+2 Setting x = 0 and x = −2 gives A = 5 and B = −2. Thus 3x + 10 1 1 dx = 5 dx + −2 dx = 5 ln |x| − 2 ln |x + 2| + C. 2 x + 2x x x+2
12. Write
A B x+1 = + . Then x + 1 = A(x − 4) + B(x + 4). 2 x − 16 x+4 x−4 Setting x = −4 and x = 4 gives A = 3/8 and B = 5/8. Thus x+1 3 1 5 1 3 5 dx = dx + dx = ln |x + 4| + ln |x − 4| + C. 2 x − 16 8 x+4 8 x−4 8 8
13. Write
1 A B = + . Then 1 = A(2x − 5) + B(2x + 5). − 25 2x + 5 2x − 5 Setting x = −5/2 and x = 5/2 gives A = −1/10 and B = 1/10. Thus 1 1 1 1 1 dx = − dx + dx 2 4x − 25 10 2x + 5 10 2x − 5
2x − 5
1 1 1
+ C. = − ln |2x + 5| + ln |2x − 5| + C = ln 20 20 20 2x + 5
14. Write
4x2
A B x = + . + 5x + 2 2x + 1 x + 2 Then x = A(x + 2) + B(2x + 1).
15. Write
2x2
Setting x = −1/2 and x = −2 gives A = −1/3 and B = 2/3. Thus 1 1 2 1 1 2 x dx = − dx + dx = − ln |2x + 1| + ln |x + 2| + C. 2 2x + 5x + 2 3 2x + 1 3 x+2 6 3 16. Write
A x+5 B C = + + . (x + 4)(x2 − 1) x+4 x−1 x+1
Then x + 5 = A(x2 − 1) + B(x + 4)(x + 1) + C(x + 4)(x − 1). Setting x = −4, x = 1, and x = −1 gives A = 1/15, B = 3/5, and C = −2/3. Thus 1 1 3 1 2 1 x+5 dx = dx + dx − dx (x + 4)(x2 − 1) 15 x+4 5 x−1 3 x+1 1 3 2 = ln |x + 4| + ln |x − 1| − ln |x + 1| + C. 15 5 3
CHAPTER 7. TECHNIQUES OF INTEGRATION
462
x2 + 2x + 6 A B C = + + . 3 x −x x x−1 x+1 Then x2 + 2x − 6 = A(x2 − 1) + B(x2 + x) + C(x2 − x).
17. Write
Setting x = 0, x = 1, and x = −1 gives A = 6, B = −3/2, and C = −7/2. Thus 2 1 3 1 7 1 x + 2x − 6 dx = 6 dx − dx − dx 3 x −x x 2 x−1 2 x+1 7 3 = 6 ln |x| − ln |x − 1| − ln |x + 1| + C. 2 2 5x2 − x + 1 A B C = + + . 3 x − 4x x x−2 x+2 Then 5x2 − x + 1 = A(x2 − 4) + B(x2 + 2x) + C(x2 − 2x).
18. Write
Setting x = 0, x = 2, and x = −2 gives A = −1/4, B = 19/8, and C = 23/8. Thus 5x2 − x + 1 1 1 19 1 23 1 dx = − dx + dx + dx 3 x − 4x 4 x 8 x−2 8 x+2 19 23 1 ln |x − 2| − ln |x + 2| + C. = − ln |x| + 4 8 8 1 A B C = + + . (x + 1)(x + 2)(x + 3) x+1 x+2 x+3 Then 1 = A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2).
19. Write
Setting x = −1, x = −2, and x = −3 gives A = 1/2, B = −1, and C = 1/2. Thus 1 1 1 1 1 1 dx = dx − dx + dx (x + 1)(x + 2)(x + 3) 2 x+1 x+2 2 x+3 1 1 = ln |x + 1| − ln |x + 2| + ln |x + 3| + C. 2 2 20. Write
(4x2
1 A B C = + + . − 1)(x + 7) 2x − 1 2x + 1 x + 7
Then 1 = A(2x + 1)(x + 7) + B(2x − 1)(x + 7) + C(4x2 − 1). Setting x = 1/2, x = −1/2, and x = −7 gives A = 1/15, B = −1/13, and C = 1/195. Thus 1 1 1 1 1 1 1 dx = dx − dx + dx 2 (4x − 1)(x + 7) 15 2x − 1 13 2x + 1 195 x+7 1 1 1 = ln |2x − 1| − ln |2x + 1| + ln |x + 7| + C. 30 26 195 4t2 + 3t − 1 A B C . = + 2+ t3 − t 2 t t t−1 Then 4t2 + 3t − 1 = A(t2 − t) + B(t − 1) + Ct2 = (A + C)t2 + (−A + B)t − B.
21. Write
Solving
A+C =4
−A+B =3
− B = −1
7.6. PARTIAL FRACTIONS
463
gives A = −2, B = 1, and C = 6. Thus 1 1 1 1 4t2 + 3t − 1 dt + dt = −2 ln |t| − + 6 ln |t − 1| + C. dt = −2 dt + 6 3 2 2 t −t t t t−1 t 2x − 11 B A C . = + 2+ x3 + 2x2 x x x+2 Then 2x − 11 = A(x2 + 2x) + B(x + 2) + Cx2 = (A + C)x2 + (2A + B)x + 2B.
22. Write
Solving
A+C =0
2A + B = 2
2B = −11
gives A = 15/4, B = −11/2, and C = −15/4. Thus 1 11 1 1 15 15 2x − 11 dx − dx dx = dx − x3 + 2x2 4 x 2 x2 4 x+2 11 15 15 ln |x| + x−1 − ln |x + 2| + C. = 4 2 4 23. Write
x3
A B C 1 = + + . 2 + 2x + x x x + 1 (x + 1)2
Then 1 = A(x + 1)2 + B(x2 + x) + Cx = (A + B)x2 + (2A + B + C)x + A. Solving
A+B =0
2A + B + C = 0
A=1
gives A = 1, B = −1, and C = −1. Thus 1 1 1 1 dx = dx − dx − dx x3 + 2x2 + x x x+1 (x + 1)2 1 = ln |x| − ln |x + 1| + + C. x+1 24. Write
t4
D t−1 A B C + = + 2+ . 3 2 + 6t + 9t t t t + 3 (t + 3)2
Then t − 1 = At(t + 3)2 + B(t + 3)2 + Ct2 (t + 3) + Dt2 = (A + C)t3 + (6A + B + 3C + D)t2 + (9A + 6B)t + 9B. Solving
A+C =0 9A + 6B = 1
6A + B + 3C + D = 0 9B = −1
gives A = 5/27, B = −1/9, C = −5/27, and D = −4/9. Thus 1 1 1 1 4 1 5 5 t−1 dt − dt − dt = dt − dt 4 3 2 2 t + 6t + 9t 27 t 9 t 27 t+3 9 (t + 3)2 5 1 4 5 = ln |t| + t−1 − ln |t + 3| + (t + 3)−1 + C. 27 9 27 9 2x − 1 2(x + 1) − 3 2 3 25. dx = dx = dx − dx 3 3 2 (x + 1) (x + 1) (x + 1) (x + 1)3 3 1 2 2 3 + +C =− − +C = x + 1 2 (x + 1)2 2(x + 1)2 x+1
CHAPTER 7. TECHNIQUES OF INTEGRATION
464 26. Write
1 B D F A C E + + = + 2+ + . x2 (x2 − 4)2 x x x − 2 (x − 2)2 x + 2 (x + 2)2
Then 1 = Ax(x2 − 4)2 + B(x2 − 4)2 + Cx2 (x − 2)(x + 2)2 + Dx2 (x + 2)2 + Ex2 (x − 2)2 (x + 2) + F x2 (x − 2)2 = (A + C + E)x5 + (B + 2C + D − 2E + F )x4 + (−8A − 4C + 4D − 4E − 4F )x3 + (−8B − 8C + 4D + 8E + 4F )x2 + 16Ax + 16B. A+C +E =0 −8A − 4C + 4D − 4E − 4F = 0 16A = 0
Solving
B + 2C + D − 2E + F = 0 −8B − 8C + 4D + 8E + 4F = 0 16B = 1
gives A = 0, B = 1/16, C = −3/128, D = 1/64, E = 3/128, and F = 1/64. Thus 1 1 1 1 1 3 1 dx = dx − dx dx − x2 (x2 − 4)2 16 x2 128 x−2 64 (x − 2)2 1 1 1 3 dx + dx + 128 x+2 64 (x + 2)2 1 1 3 = − x−1 − ln |x − 2| − (x − 2)−1 16 128 64 1 3 ln |x + 2| − (x + 2)−1 + C. + 128 64 27. Write
1 B D A C + + = + . (x2 + 6x + 5)2 x + 1 (x + 1)2 x + 5 (x + 5)2
Then 1 = A(x + 1)(x + 5)2 + B(x + 5)2 + C(x + 5)(x + 1)2 + D(x + 1)2 = (A + C)x3 + (11A + B + 7C + D)x2 + (35A + 10B + 11C + 2D)x + 25A + 25B + 5C + D. A+C =0 35A + 10B + 11C + 2D = 0
Solving
11A + B + 7C + D = 0 25A + 25B + 5C + D = 1
gives A = −1/32, B = 1/16, C = 1/32, and D = 1/16. Thus 1 1 1 1 1 1 1 dx + dx dx = − dx + 2 2 2 (x + 6x + 5) 32 x+1 16 (x + 1) 32 x+5 1 1 + dx 16 (x + 5)2 1 1 1 1 1 1 ln |x + 5| − = − ln |x + 1| − + + C. 32 16 x + 1 32 16 x + 5 28. Write
(x2
A B C D 1 = + + + . 2 − x − 6)(x − 2x − 8) x − 4 x − 3 x + 2 (x + 2)2
Then 1 = A(x − 3)(x + 2)2 + B(x − 4)(x + 2)2 + C(x − 4)(x − 3)(x + 2) + D(x − 4)(x − 3) = (A + B + C)x3 + (A − 5C + D)x2 + (−8A − 12B − 2C − 7D)x + (−12A − 16B + 24C + 12D).
7.6. PARTIAL FRACTIONS Solving
465
A+B+C =0 −8A − 12B − 2C − 7D = 0
A − 5C + D = 0 −12A − 16B + 24C + 12D = 1
gives A = 1/36, B = −1/25, C = 11/900, and D = 1/30. (Note that A and B can be easily obtained by substituting x = 4 and x = 3, respectively, in the initial equation.) Thus 1 1 1 1 11 1 1 dx = dx − dx + dx (x2 − x − 6)(x2 − 2x − 8) 36 x−4 25 x−3 900 x+2 1 1 + dx 30 (x + 2)2 1 1 11 = ln |x − 4| − ln |x − 3| + ln |x + 2| 36 25 900 1 − (x + 2)−1 + C. 30 B A C D E x4 + 2x2 − x + 9 . = + 2+ 3+ 4+ x5 + 2x4 x x x x x+2 Then x4 + 2x2 − x + 9 = Ax3 (x + 2) + Bx2 (x + 2) + Cx(x + 2) + D(x + 2) + Ex4
29. Write
= (A + E)x4 + (2A + B)x3 + (2B + C)x2 + (2C + D)x + 2D. Solving
A+E =1 2C + D = −1
2A + B = 0 2D = 9
2B + C = 2
gives A = −19/16, B = 19/8, C = −11/4, D = 9/2, and E = 35/16. Thus
1 19 1 1 11 dx + dx − dx 2 x 8 x 4 x3 1 1 9 35 dx + dx + 2 x4 16 x+2 19 19 1 3 1 35 11 1 = − ln |x| − ln |x + 2| + C. − + + 16 8 x 8 x2 2 x3 16
19 x4 + 2x2 − x + 9 dx = − 5 4 x + 2x 16
30. Write
B C E 5x − 1 A D + + = + + . 2 2 2 x(x − 3) (x + 2) x x − 3 (x − 3) x + 2 (x + 2)2
Then 5x − 1 = A(x − 3)2 (x + 2)2 + Bx(x − 3)(x + 2)2 + Cx(x + 2)2 + Dx(x − 3)2 (x + 2) + Ex(x − 3)2 = (A + B + D)x4 + (−2A + B + C + 4D + E)x3 + (11A − 8B + 4C − 3D − 6E)x2 + (12A − 12B + 4C + 18D + 9E)x + 36A. Solving
A+B+D =0 11A − 8B + 4C − 3D − 6E = 0 36A = −1
−2A + B + C − 4D + E = 0 12A − 12B + 4C + 18D + 9E = 5
CHAPTER 7. TECHNIQUES OF INTEGRATION
466
gives A = −1/36, B = −79/1125, C = 14/75, D = 49/500, and E = 11/50. Thus 1 79 1 14 1 5x − 1 1 dx − dx + dx = − dx x(x − 3)2 (x + 2)2 36 x 1125 x−3 75 (x − 3)2 1 11 1 49 dx + dx + 500 x+2 50 (x + 2)2 1 79 14 49 = − ln |x| − ln |x − 3| − (x − 3)−1 + ln |x + 2| 36 1125 75 500 11 − (x + 2)−1 + C. 50 31. Write
A Bx + C x−1 = + 2 . 2 x(x + 1) x x +1
Then x − 1 = A(x2 + 1) + (Bx + C)x = (A + B)x2 + Cx + A. Solving
A+B =0
A = −1
C=1
gives A = −1, B = 1, and C = 1. Thus 1 x 1 x−1 dx = − dx + dx + dx x(x2 + 1) x x2 + 1 x2 + 1 1 = − ln |x| + ln(x2 + 1) + tan−1 x + C. 2 32. Write
A Bx + C 1 = + 2 . 2 (x − 1)(x + 3) x−1 x +3
Then 1 = A(x2 + 3) + (Bx + C)(x − 1) = (A + B)x2 + (−B + C)x + (3A − C). Solving
A+B =0
−B+C =0
3A − C = 1
gives A = 1/4, B = −1/4, and C = −1/4. Thus 1 1 1 x 1 1 1 dx = dx − dx − dx (x − 1)(x2 + 3) 4 x−1 4 x2 + 3 4 x2 + 3 1 x 1 1 = ln |x − 1| − ln(x2 + 3) − √ tan−1 √ + C. 4 8 4 3 3 33. Write
A B x Cx + D = + . + 2 (x + 1)2 (x2 + 1) x + 1 (x + 1)2 x +1
Then x = A(x + 1)(x2 + 1) + B(x2 + 1) + (Cx + D)(x + 1)2 = (A + C)x3 + (A + B + 2C + D)x2 + (A + C + 2D)x + (A + B + D). Solving
A+C =0 A + C + 2D = 1
A + B + 2C + D = 0 A+B+D =0
gives A = 0, B = −1/2, C = 0, and D = 1/2. Thus 1 1 1 1 1 1 x 1 dx = − dx = dx + + tan−1 x + C. (x + 1)2 (x2 + 1) 2 (x + 1)2 2 x2 + 1 2 x+1 2
7.6. PARTIAL FRACTIONS 34. Write
467
x2 A B C Dx + E = + . + + 2 3 2 2 3 (x − 1) (x + 4) x − 1 (x − 1) (x − 1) x +4
Then x2 = A(x − 1)2 (x2 + 4) + B(x − 1)(x2 + 4) + C(x2 + 4) + (Dx + E)(x − 1)3 = (A + D)x4 + (−2A + B − 3D + E)x3 + (5A − B + C + 3D − 3E)x2 + (−8A + 4B − D + 3E)x + (4A − 4B + 4C − E). Solving
A+D =0 5A − B + C + 3D − 3E = 1 4A − 4B + 4C − E = 0
−2A + B − 3D + E = 0 −8A + 4B − D + 3E = 0
gives A = 4/125, B = 8/25, C = 1/5, D = −4/125, and E = −44/125. Thus
35. Write
1 8 1 1 1 dx + dx + dx x−1 25 (x − 1)2 5 (x − 1)3 x 44 1 4 dx − dx − 2 2 125 x +4 125 x +4 8 1 4 ln |x − 1| − (x − 1)−1 − (x − 1)−2 = 125 25 10 x 2 22 ln(x2 + 4) − tan−1 + C. − 125 125 2
x2 4 dx = (x − 1)3 (x2 + 4) 125
x4
Ax + B Cx + D 1 = 2 + 2 . 2 + 5x + 4 x +1 x +4
Then 1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1) = (A + C)x3 + (B + D)x2 + (4A + C)x + (4B + D). Solving
A+C =0
B+D =0
4A + C = 0
4B + D = 1
gives A = 0, B = 1/3, C = 0, and D = −1/3. Thus 1 1 1 1 1 x 1 1 dx = dx − dx = tan−1 x − tan−1 + C. x4 + 5x2 + 4 3 x2 + 1 3 x2 + 4 3 6 2 36. Write
1 Ax + B Cx + D = 2 + 2 . x4 + 13x2 + 36 x +9 x +4
Then 1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 9) = (A + C)x3 + (B + D)x2 + (4A + 9C)x + (4B + 9D). Solving
A+C =0
B+D =0
4A + 9C = 0
4B + 9D = 1
gives A = 0, B = −1/5, C = 0, and D = 1/5. Thus x x 1 1 1 1 1 1 1 dx = − dx + dx = − tan−1 + tan−1 + C. x4 + 13x2 + 36 5 x2 + 9 5 x2 + 4 15 3 10 2
CHAPTER 7. TECHNIQUES OF INTEGRATION
468
1 A Bx + C = + . x3 − 1 x − 1 x2 + x + 1 Then 1 = A(x2 + x + 1) + (Bx + C)(x − 1) = (A + B)x2 + (A − B + C)x + (A − C).
37. Write
Solving
A+B =0
A−B+C =0
A−C =1
gives A = 1/3, B = −1/3, and C = −2/3. Thus 1 1 1 2x + 4 1 dx = dx − dx 3 2 x −1 3 x−1 6 x +x+1 1 1 2x + 1 1 1 = ln |x − 1| − dx − dx 3 6 x2 + x + 1 2 x2 + x + 1 1 1 1 1 = ln |x − 1| − ln |x2 + x + 1| − dx 3 6 2 (x + 1/2)2 + 3/4 1 1 2x + 1 1 = ln |x − 1| − ln |x2 + x + 1| − √ tan−1 √ + C. 3 6 3 3 81 A B Cx + D = + + . x4 + 27x x x + 3 x2 − 3x + 9 Then 81 = A(x3 + 27) + B(x3 − 3x2 + 9x) + (Cx + D)(x2 + 3x)
38. Write
= (A + B + C)x3 + (−3B + 3C + D)x2 + (9B + 3D)x + 27A. Solving
A+B+C =0 9B + 3D = 0
−3B + 3C + D = 0 27A = 81
gives A = 3, B = −1, C = −2, and D = 3. Thus 1 1 2x − 3 81 dx = 3 dx − dx − dx x4 + 27x x x+3 x2 − 3x + 9
x3
+ C.
= 3 ln |x| − ln |x + 3| − ln |x − 3x + 9| + C = ln 3 x + 27
2
39. Write
3x2 − x + 1 A Bx + C = + . (x + 1)(x2 + 2x + 2) x + 1 x2 + 2x + 2
Then 3x2 − x + 1 = A(x2 + 2x + 2) + (Bx + C)(x + 1) = (A + B)x2 + (2A + B + C)x + (2A + C). Solving
A+B =3
2A + B + C = −1
2A + C = 1
gives A = 5, B = −2, and C = −9. Thus 1 2x + 9 3x2 − x + 1 dx = 5 dx − dx (x + 1)(x2 + 2x + 2) x+1 x2 + 2x + 2 7 2x + 2 dx − dx = 5 ln |x + 1| − 2 2 x + 2x + 2 x + 2x + 2 7 = 5 ln |x + 1| − ln |x2 + 2x + 2| − dx (x + 1)2 + 1 = 5 ln |x + 1| − ln |x2 + 2x + 2| − 7 tan−1 (x + 1) + C.
7.6. PARTIAL FRACTIONS 40. Write
469
4x + 12 A Bx + C = + . (x − 2)(x2 + 4x + 8) x − 2 x2 + 4x + 8
Then 4x + 12 = A(x2 + 4x + 8) + (Bx + C)(x − 2) = (A + B)x2 + (4A − 2B + C)x + (8A − 2C). Solving
4A − 2B + C = 4
A+B =0
8A − 2C = 12
gives A = 1, B = −1, and C = −2. Thus 1 x+2 4x + 12 dx = dx − dx (x − 2)(x2 + 4x + 8) x−2 x2 + 4x + 8 2x + 4 1 dx = ln |x − 2| − 2 2 x + 4x + 8 1 = ln |x − 2| − ln |x2 + 4x + 8| + C. 2 41. Write
Cx + D x2 − x + 4 Ax + B + 2 = 2 . (x2 + 4)2 x +4 (x + 4)2
Then x2 − x + 4 = (Ax + B)(x2 + 4) + Cx + D = Ax3 + Bx2 + (4A + C)x + (4B + D). Solving
A=0
4A + C = −1
B=1
4B + D = 4
gives A = 0, B = 1, C = −1, and D = 0. Thus 2 x 1 1 1 1 x −x+4 −1 x dx − tan + dx = dx = +C (x2 + 4)2 x2 + 4 (x2 + 4)2 2 2 2 x2 + 4 42. Write
B Fx + G A C Dx + E 1 + 2 = + 2+ 3+ 2 . x3 (x2 + 1)2 x x x x +1 (x + 1)2
Then 1 = Ax2 (x2 + 1)2 + Bx(x2 + 1)2 + C(x2 + 1)2 + (Dx + E)x3 (x2 + 1) + (F x + G)x3 = (A + D)x6 + (B + E)x5 + (2A + C + D + F )x4 + (2B + E + G)x3 + (A + 2C)x2 + Bx + C. Solving
A+D =0 2B + E + G = 0
B+E =0 A + 2C = 0
2A + C + D + F = 0 B=0 C=1
gives A = −2, B = 0, C = 1, D = 2, E = 0, F = 1, and G = 0. Thus 1 1 2x 1 2x 1 dx + dx + dx = −2 dx + dx 3 2 2 3 2 2 x (x + 1) x x x +1 2 (x + 1)2 1 1 = −2 ln |x| − x−2 + ln(x2 + 1) − (x2 + 1)−1 + C. 2 2 43. For this and possibly later problems, we will encounter cos2 θ dθ. Using Example 12 of Section 1.4, 5.2 in the text, we have 1 1 1 1 cos2 θ dθ = θ + sin 2θ + C = θ + sin θ cos θ + C. 2 4 2 2
CHAPTER 7. TECHNIQUES OF INTEGRATION
470 Write
Ax + B Cx + D x3 − 2x2 + x − 3 = 2 + 2 . 4 2 x + 8x + 16 x +4 (x + 4)2
4 + x2
Then x3 − 2x2 + x − 3 = (Ax + B)(x2 + 4) + Cx + D
θ
= Ax3 + Bx2 + (4A + C)x + (4B + D). Solving
A=1
B = −2
4A + C = 1
x
2
4B + D = −3
gives A = 1, B = −2, C = −3, and D = 5. Thus 3 x 1 x x − 2x2 + x − 3 dx = dx − 2 dx − 3 dx 4 2 2 2 2 x + 8x + 16 x +4 x +4 (x + 4)2 1 dx x = 2 tan θ, dx = 2 sec2 θ dθ +5 2 (x + 4)2 3 1 1 2 sec2 θ 2 −1 x + = ln(x + 4) − tan dθ + 5 2 2 2 x2 + 4 (4 tan2 θ + 4)2 x 3 5 1 + cos2 θ dθ = ln(x2 + 4) − tan−1 + 2 2 2(x2 + 4) 8 1 3 5 5 x = ln(x2 + 4) − tan−1 + + θ+ sin θ cos θ + C 2 2 2(x2 + 4) 16 16 1 x x 3 5 = ln(x2 + 4) − tan−1 + + tan−1 2 2 2 2(x + 4) 16 2 x 5 + +C 8 x2 + 4 5x + 12 1 x 11 tan−1 + + C. = ln(x2 + 4) − 2 6 2 8(x2 + 4) 44. Write
Cx + D x2 Ax + B + 2 = 2 . 2 2 (x + 3) x +3 (x + 3)2
3 + x2
Then x2 = (Ax + B)(x2 + 3) + Cx + D
θ
= Ax3 + Bx2 + (3A + C)x + (3B + D). Solving
A=0
B=1
3A + C = 0
√
x 3
3B + D = 0
gives A = 0, B = 1, C = 0, and D = −3. Thus √ √ 1 x2 1 dx − 3 dx = dx x = 3 tan θ, dx = 3 sec2 θ dθ 2 2 2 2 2 (x + 3) x +3 (x + 3) √ x 3 sec2 θ 1 1 1 −1 x √ √ √ tan − cos2 θ dθ = √ tan−1 √ − 3 dθ = (3 tan2 θ + 3)2 3 3 3 3 3 x 1 1 1 = √ tan−1 √ − √ θ − √ sin θ cos θ + C 3 3 2 3 2 3 x x 1 1 1 −1 −1 x = √ tan √ − √ tan √ − +C 3 3 2 3 3 2 x2 + 3 x x 1 +C = √ tan−1 √ − 2 2x +6 2 3 3
7.6. PARTIAL FRACTIONS
471
x4 + 3x2 + 4 10x + 2 10(x + 1) − 8 = x2 − 2x + 6 − = x2 − 2x + 6 − . 2 2 (x + 1) (x + 1) (x + 1)2 4 1 x + 3x2 + 4 1 2 dx + 8 Then dx = (x − 2x + 6) dx − 10 dx 2 (x + 1) x+1 (x + 1)2 1 8 = x3 − x2 + 6x − 10 ln |x + 1| − + C. 3 x+1 A 9x B C D x5 − 10x3 = x − = x − + + + 46. Write 4 . x − 10x2 + 9 x4 − 10x2 + 9 x−1 x+1 x−3 x+3
45. Write
Then 9x = A(x + 1)(x2 − 9) + B(x − 1)(x2 − 9) + C(x + 3)(x2 − 1) + D(x − 3)(x2 − 1). Setting x = 1, x = −1, x = 3, and x D = 9/16. Thus x5 − 10x3 dx = x dx − − x4 − 10x2 + 9 9 + 16
= −3 gives A = −9/16, B = −9/16, C = 9/16, and 9 16
1 9 dx − x−1 16
1 9 dx + x+1 16
1 dx x−3
1 dx x+3
1 9 9 9 9 ln |x − 1| + ln |x + 1| − ln |x − 3| − ln |x + 3| + C + x2 16 16 16 16
2
x − 1
1 9
+ C. = x2 + ln
2 2 16 x − 9
=
A B 1 = + . x2 − 6x + 5 x−1 x−5 Then 1 = A(x − 5) + B(x − 1). Setting x = 1 and x = 5 gives A = −1/4 and B = 1/4. Thus 4 4 4 1 4 1 1 4 1 1 1 1 dx = − dx + dx = − ln |x − 1| + ln |x − 5| 2 4 2 x−1 4 2 x−5 4 4 2 x − 6x + 5 2 2
4
x − 5
1 1 1 1 = ln
= ln − ln 3 = − ln 3. 4 x − 1 2 4 3 2
47. Write
1 A B = + . −4 x−2 x+2 Then 1 = A(x + 2) + B(x − 2). Setting x = 2 and x = −2 gives A = 1/4 and B = −1/4. 1 1 1 1 1 1 1 1 1 1 1 1 dx = dx − dx = ln |x − 2| ln |x + 2| Thus − 2 4 0 x−2 4 0 x+2 4 4 0 x −4 0 0
1 1 1
x − 2
1 1 = ln
= ln − ln 1 = − ln 3. 4 x + 2 0 4 3 4 2 2 2 2 2(x + 3) − 7 2 1 2x − 1 49. dx − 7 dx = dx = dx 2 2 2 (x + 3) 0 (x + 3) 0 0 x+3 0 (x + 3) 2 2 7 7 7 5 14 − = 2(ln 5 − ln 3) + = 2 ln |x + 3| + = 2 ln − x+3 0 5 3 3 15 0
48. Write
x2
CHAPTER 7. TECHNIQUES OF INTEGRATION
472 50. Write
2x + 6 B C A + = + . x(x + 1)2 x x + 1 (x + 1)2
Then 2x + 6 = A(x + 1)2 + B(x2 + x) + Cx = (A + B)x2 + (2A + B + C)x + A. Solving
A+B =0
2A + B + C = 2
A=6
gives A = 6, B = −6, and C = −4. Thus 5 5 5 5 2x + 6 1 1 1 dx − 6 dx − 4 = 6 dx 2 2 1 x(x + 1) 1 x 1 x+1 1 (x + 1)
5 5 5 5
x 5 4
+ 4 = 6 ln
= 6 ln |x| − 6 ln |x + 1| + x+1 1 x + 1 1 x+1 1 1 1 5 1 4 4 5 4 − ln − =6 + = 6 ln − . 6 2 6 2 3 3 1 A Bx + C = + 2 . + + 2x + 2 x+1 x +2 Then 1 = A(x2 + 2) + (Bx + C)(x + 1) = (A + B)x2 + (B + C)x + (2A + C).
51. Write
x3
Solving
x2
A+B =0
B+C =0
2A + C = 1
gives A = 1/3, B = −1/3, and C = 1/3. Thus 1 1 1 1 1 1 2x 1 1 1 1 dx = dx − dx + dx 3 2 3 0 x+1 6 0 x2 + 2 3 0 x2 + 2 0 x + x + 2x + 2 1 1 1 1 x 1 1 = ln |x + 1| + ln(x2 + 2) + √ tan−1 √ 3 6 3 2 2 0 0 0 1 1 1 1 = ln 2 − (ln 3 − ln 2) + √ tan−1 √ 3 6 3 2 2 1 1 8 1 = ln + √ tan−1 √ . 6 3 3 2 2
1
52. 0
x2 dx = x4 + 8x2 + 16
1 0
x2 + 4 − 4 dx = (x2 + 4)2
1 0
1 dx − 4 x2 + 4
1 0
1 dx (x2 + 4)2
2
= = = =
x = 2 tan θ, dx = 2 sec θ dθ 1 tan−1 1/2 2 sec2 θ 1 −1 x tan −4 dθ 2 2 0 (4 tan2 θ + 4)2 0 tan−1 1/2 √ 1 1 1 5 tan−1 − cos2 θ dθ 2 2 2 0 tan−1 1/2 2 1 1 1 1 −1 1 tan − θ + sin θ cos θ 2 2 2 2 2 0 2 1 1 1 1 1 1 1 1 1 √ √ tan−1 − tan−1 − = tan−1 − 2 2 4 2 4 4 2 10 5 5
1
7.6. PARTIAL FRACTIONS
1
53. −1
473
1 1 1 4x3 + 10x 1 2x3 + 5x 4 2 dx = dx = ln |x + 5x + 6| x4 + 5x2 + 6 2 −1 x4 + 5x2 + 6 2 −1 1 = (ln 12 − ln 12) = 0 2
54. Write
B A C Dx + E 1 . = + 2+ 3+ 2 x5 + 4x4 + 5x3 x x x x + 4x + 5
Then 1 = Ax2 (x2 + 4x + 5) + Bx(x2 + 4x + 5) + C(x2 + 4x + 5) + (Dx + E)x3 = (A + D)x4 + (4A + B + E)x3 + (5A + 4B + C)x2 + (5B + 4C)x + 5C. Solving
A+D =0 5B + 4C = 0
4A + B + E = 0 5C = 1
5A + 4B + C = 0
gives A = 11/125, B = −4/25, C = 1/5, D = −11/125, and E = −24/125. Thus
2 1
√ 55.
2 4 1 1 1 2 1 dx − dx + dx 25 1 x2 5 1 x3 1 x 2 2 2 2x + 4 1 11 dx − dx − 2 250 1 x + 4x + 5 125 1 (x + 2)2 + 1 2 2 2 2 1 11 4 1 1 11 2 ln |x| + ln |x = − − + 4x + 5| 125 25 x 1 10 x2 1 250 1 1 2 2 tan−1 (x + 1) − 125 1 4 1 1 1 11 11 ln 2 + −1 − −1 − (ln 17 + ln 10) = 125 25 2 10 4 250 2 (tan−1 4 − tan−1 3) − 125 40 1 2 11 ln − − (tan−1 4 − tan−1 3). = 250 17 200 125
1 11 dx = 5 4 3 x + 4x + 5x 125
1 − x2 dx = x3
Write
2
√
1 − x2 x dx u2 = 1 − x2 , 2u du = −2x dx x4 u u2 = (−u du) = − du (1 − u2 )2 (1 − u2 )2
B D A C u2 + + = + . 2 2 2 (1 − u ) 1 − u (1 − u) 1 + u (1 + u)2
Then u2 = A(1 − u)(1 + u)2 + B(1 + u2 ) + C(1 + u)(1 − u)2 + D(1 − u)2 = (A + B + C + D) + (A + 2B − C − 2D)u + (−A + B − C + D)u2 + (−A + C)u3 . Solving
A+B+C +D =0
A + 2B − C − 2D = 0
−A + B − C + D = 1
−A + C = 0
CHAPTER 7. TECHNIQUES OF INTEGRATION
474
gives A = −1/4, B = 1/4, C = −1/4, and D = 1/4. Thus √
1 − x2 u2 dx = − du x3 (1 − u2 )2 1 1 1 1 1 1 1 1 du − du − du + du = 4 1−u 4 (1 − u)2 4 1+u 4 (1 + u)2 1 1 1 1 1 1 = − ln |1 − u| − + ln |1 + u| + +C 4 4 1−u 4 4 1+u
√ √
1 − x2 u 1
1 + 1 − x2
1 1
1 + u
1 √ − + C. + C = ln
= ln
− 4 1 − u 2 1 − u2 4 1 − 1 − x2 2 x2
56.
Write
x−1 dx x+1
1 + u2 x−1 4u , x= , dx = 2 du 2 x+1 1−u (u − 1)2 4u 4u2 = |u| 2 du = du (u − 1)2 (u2 − 1)2 u2 =
4u2 B D A C = + + + . Then (u2 − 1)2 u − 1 (u − 1)2 u + 1 (u + 1)2
4u2 = A(u − 1)(u + 1)2 + B(u + 1)2 + C(u − 1)2 (u + 1) + D(u − 1)2 = (A + C)u3 + (A + B − C + D)u2 + (−A + 2B − C − 2D)u + (−A + B + C + D).
Solving
A+C =0 −A + 2B − C − 2D = 0
A+B−C +D =4 −A + B + C + D = 0
gives A = 1, B = 1, C = −1, and D = 1. Thus
x−1 dx = x+1
1 1 du + du u+1 (u + 1)2 1 1 − ln |u + 1| − +C u−1 u+1
x−1 − 1
2
x+1
2 − 1 − x
+
− x x2 − 1 + C. + C = ln
x−1
+ 1 x + 1 − 1
1 du + u−1
= ln |u − 1| −
x−1
x+1 = ln
x−1
x+1
1 du − (u − 1)2
7.6. PARTIAL FRACTIONS √ 3 57.
x+1 dx x
475
u (3u2 du) u3 = x + 1, 3u2 du = dx = 3 u −1 3 1 1 u −1+1 du = 3 du + 3 du = 3u + du =3 3 3 3 u −1 u −1 u −1 √ 2u + 1 1 +C = 3u + ln |u − 1| − ln |u2 + u + 1| − 3 tan−1 √ 2 3
√ √ √ 1
= 3 3 x + 1 + ln | 3 x + 1 − 1| − ln 3 (x + 1)2 + 3 x + 1 + 1
2 √ 3 √ −1 2 x + 1 + 1 √ − 3 tan + C. 3
1 √ x(1 + 3 x)2
x = u6 , dx = 6u5 du 6u5 6u2 = du = du u = tan θ, du = sec2 θ dθ u3 (1 + u2 )2 (1 + u2 )2 tan2 θ tan2 θ 2 dθ = 6 sin2 θ dθ =6 sec θ dθ = 6 sec2 θ (1 + tan2 θ)2 3 = 3 (1 − cos 2θ) dθ = 3θ − sin 2θ + C = 3θ − 3 sin θ cos θ + C 2 u 3x1/6 −1 1/6 = 3 tan−1 u − 3 + C = 3 tan x − +C 1 + u2 1 + x1/3 A B 1 1 = + . 59. Write 2 x + 2x − 3 x+3 x−1
58.
√
2 4 Then 1 = A(x − 1) + B(x + 3). Setting x = −3 and x = 1 gives A = −1/4 and B = 1/4. Thus 4 4 1 4 1 1 4 1 1 1 Area = − dx + dx = − ln |x + 3| + ln |x − 1| 4 2 x+3 4 2 x−1 4 4 2 2
4 1 1
x − 1
1 3 1 15 ≈ 0.1905. = ln
= ln − ln = ln 4 x + 3 2 4 7 5 4 7
60. Write
Ax + B Cx + D x3 = 2 + 2 . (x2 + 1)(x2 + 2) x +1 x +2
1
Then x3 = (Ax + B)(x2 + 2) + (Cx + D)(x2 + 1) 3
2
4
2
= (A + C)x + (B + D)x + (2A + C)x + (2B + D). Solving
A+C =1
B+D =0
2A + C = 0
2B + D = 0
gives A = −1, B = 0, C = 2, and D = 0. Thus 4 4 4 4 x 2x 1 2 2 dx + dx = − ln(x Area = − + 1) + ln(x + 2) 2 2 2 0 0 x +1 0 x +2 0 4 9 x2 + 2 18 = ln √ = ln √ − ln 2 = ln √ ≈ 0.7806 17 17 x2 + 1 0
CHAPTER 7. TECHNIQUES OF INTEGRATION
476
A B x = + . Then x = A(x + 3) + B(x + 2). (x + 2)(x + 3) x+2 x+3 Setting x = −2 and x = −3 gives A = −2 and B = 3. Thus 0 1 −2 −2 3 3 Area = − + + dx + dx x+2 x+3 x+2 x+3 −1 0 1 0 + (−2 ln |x + 2| + 3 ln |x + 3|) = −(−2 ln |x + 2| + 3 ln |x + 3|)
1
61. Write
−1
-1 1
1
-1
0
= −[(−2 ln 2 + 3 ln 3) − (−2 ln 1 + 3 ln 2)] + [(−2 ln 3 + 3 ln 4) − (−2 ln 2 + 3 ln 3)] 8192 ≈ 0.2220. = 7 ln 2 − 8 ln 3 + 3 ln 4 = ln 6561 A 24 Bx + C 3x3 2 =3+ 3 = 3 + 24 + 2 62. Write 3 . x −8 x −8 x − 2 x + 2x + 4 1
Then 1 = A(x2 + 2x + 4) + (Bx + C)(x − 2) = (A + B)x2 + (2A − 2B + C)x + (4A − 2C). Solving
A+B =0
2A − 2B + C = 0
-2
4A − 2C = 1
-1
1 -1
gives A = 1/12, B = −1/12, and C = −1/3. Thus 1/12 x/12 1/3 3x3 dx = 3 + 24 − − dx x3 − 8 x − 2 x2 + 2x + 4 x2 + 2x + 4 2x + 8 2 − 2 dx = 3+ x − 2 x + 2x + 4 1 1 1 dx − (2x + 2) dx − 6 dx = 3 dx + 2 2 2 x−2 x + 2x + 4 x + 2x + 4 1 = 3x + 2 ln |x − 2| − ln |x2 + 2x + 4| − 6 dx (x + 1)2 + 3
2 √
x − 4x + 4
+1
− 2 3 tan−1 x√ +C = 3x + ln
2 x + 2x + 4
3
2
√
x − 4x + 4
+1 3x3
− 2 3 tan−1 x√ dx = 3x + ln Let g(x) = . Then
3 2 x −8 x + 2x + 4 3 1 0 0 1 3x3 3x3 dx − dx = g(x) − g(x) = [g(0) − g(−2)] − [g(1) − g(0)] Area = 3 3 −2 0 −2 x − 8 0 x −8 √ √ 1 −1 = 0 + ln 1 − 2 3 tan−1 √ − −6 + ln 4 − 2 3 tan−1 √ 3 3 √ √ 1 2 1 − 3 + ln − 2 3 tan−1 √ − 0 + ln 1 − 2 3 tan−1 √ 7 3 3 √ √ √ √ 3π 3π 2 3π + 6 − ln 4 − − 3 − ln 7 − 2 3 tan−1 √ + = − 3 3 3 3 √ √ 7 2 = 3 + ln − 3π + 2 3 tan−1 √ ≈ 1.0872. 4 3
7.6. PARTIAL FRACTIONS
477
3
4 dx 2 (x + 1)2 x 1 B D A C 4 + = + 2+ . Write 2 x (x + 1)2 x x x + 1 (x + 1)2
63. V = π
1
3
Then 4 = Ax(x + 1)2 + B(x + 1)2 + Cx2 (x + 1) + Dx2 = (A + C)x3 + (2A + B + C + D)x2 + (A + 2B)x + B. Solving
A+C =0
2A + B + C + D = 0
A + 2B = 0
B=4
gives A = −8, B = 4, C = 8, and D = 4. Thus 3 3 3 3 1 1 1 1 dx + 4 dx + 4 V = π −8 dx + 8 dx 2 2 1 x 1 x 1 x+1 1 (x + 1)
3 3
x + 1
8x + 4
4 4
− = π −8 ln |x| − + 8 ln |x + 1| − = π 8 ln
x x+1 1 x x(x + 1) 1 4 7 2 11π ≈ 1.3287. = π 8 ln − − (8 ln 2 − 6) = 8π ln + 3 3 3 3
2
1 dx 0 (x + 1)(x + 4) A B 1 = + . Write (x + 1)(x + 4) x+1 x+4
64. V = π
1
2
Then 1 = A(x + 4) + B(x + 1). Setting x = −1 and x = −4 gives A = 1/3 and B = −1/3. Thus 2 2 1 1 1 2 1 1 1 V =π dx − dx = π ln |x + 1| − ln |x + 4| 3 0 x+1 3 0 x+4 3 3 0
2
x + 1
1 π π 1 π = ln
= ln − ln = ln 2 ≈ 0.7259. 3 x + 4 0 3 2 4 3
1 4x x+1−1 dx = 8π dx 2 2 0 (x + 1) 0 (x + 1) 1 1 1 1 = 8π dx − dx 2 0 x+1 0 (x + 1) 1 1 1 = 8π ln |x + 1| + = 8π ln 2 + − (ln 1 + 1) x+1 0 2 1
65. V = 2π
4
= 8π ln 2 − 4π ≈ 4.8543
1
CHAPTER 7. TECHNIQUES OF INTEGRATION
478
1
66. V = 2π 0
Write
(x2
(x2
8x dx + 1)(x2 + 4)
2
Ax + B Cx + D 8x = 2 + 2 . 2 + 1)(x + 4) x +1 x +4
Then 8x = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)
1
= (A + C)x3 + (B + D)x2 + (4A + C)x + (4B + D). Solving
A+C =0
B+D =0
4A + C = 8
4B + D = 0
gives A = 8/3, B = 0, C = −8/3, and D = 0. Thus 1 1
8 4 8 1 x x 4 2 2
dx − dx = 2π ln(x ln(x + 1) − + 4)
2 2 3 0 x +1 3 0 x +4 3 3 0 1 1 8π x2 + 1 8π 2 8π 8 ln 2 ln ≈ 3.9375. = = ln − ln = 3 x +4 0 3 5 4 3 5
V = 2π
67.
cos x u = sin x, du = cos x dx dx sin2 x + 3 sin x + 2 A B 1 = + . Write (u + 1)(u + 2) u+1 u+2
=
1 du u2 + 3u + 2
Then 1 = A(u + 2) + B(u + 1). Setting u = −1 and u = −2 gives A = 1 and B = −1. Thus
68.
1 1 du − du = ln |u + 1| − ln |u + 2| + C u+1 u+2
u + 1
+ C = ln sin x + 1 + C. = ln
u+2 sin x + 2
cos x dx = sin2 x + 3 sin x + 2
sin x dx cos2 x − cos3 x
Write
u = cos x, du = − sin x dx
=
−1 du = u2 − u3
1 du u2 (u − 1)
A B C 1 = + 2+ . u2 (u − 1) u u u−1
Then 1 = A(u2 − u) + B(u − 1) + Cu2 = (A + C)u2 + (−A + B)u − B. Solving
A+C =0
−A+B =0
−B =1
gives A = −1, B = −1, and C = 1. Thus 1 1 1 1 sin x dx = − du − du = − ln |u| + + ln |u − 1| + C du + cos2 x − cos3 x u u2 u−1 u
u − 1 1
cos x − 1
+ 1 +C + + C = ln
= ln
u u cos x cos x = ln |1 − sec x| + sec x + C.
7.6. PARTIAL FRACTIONS 69.
479
et dt (et + 1)2 (et − 2)
Write
u = et , du = et dt
=
1 du (u + 1)2 (u − 2)
1 A B C = + . + 2 2 (u + 1) (u − 2) u + 1 (u + 1) u−2
Then 1 = A(u + 1)(u − 2) + B(u − 2) + C(u + 1)2 = (A + C)u2 + (−A + B + 2C)u + (−2A − 2B + C). − A + B + 2C = 0
A+C =0
Solving
− 2A − 2B + C = 1
gives A = −1/9, B = −1/3, and C = 1/9. Thus
70.
1 1 1 1 1 et 1 dt = − du − du du + (et + 1)2 (et − 2) 9 u+1 3 (u + 1)2 9 u−2 1 1 1 1 = − ln |u + 1| + + ln |u − 2| + C 9 3 u+1 9
1 1 et − 2
+ + C. = ln
t 9 e + 1 3(et + 1)
e2t dt = (et + 1)3
u−1 du u3 1 1 1 +C − t = (u−2 − u−3 ) du = −u−1 + u−2 + C = 2 2(et + 1)2 e +1 et et dt (et + 1)3
71. y = ex ln 2 L= 1 + e2x dx =
0
√
√
2
5
u2 du = 2 u −1
t
t
u = e + 1, du = e dt
u2 = 1 + e2x , 2u du = 2e2x dx, dx =
√ √
5
1+
2
1 u2 − 1
=
u u du du = 2 e2x u −1
du
A B 1 = + . u2 − 1 u−1 u+1 Then 1 = A(u + 1) + B(u − 1). Setting u = 1 and u = −1 gives A = 1/2 and B = −1/2. √5 √5 √5 1 1 1 1 du − du Thus L = √ du + √ √ u u − 1 2 u + 1 2 2 2 √
√5 5 1 1
u − 1
1 = u + ln |u − 1| − ln |u + 1| √ = u + ln
2 2 2 u + 1 √2 2 √ √ √ √ 5−1 2−1 1 1 − ≈ 1.2220. = 5 + ln √ 2 + ln √ 2 2 5+1 2+1
Write
72. (a)
x3 x3 1 1 dx = dx = (4x3 dx). Partial fraction decompo(x2 − 1)(x2 + 1) x4 − 1 4 x4 − 1 sition is unnecessary because the substitution u = x4 − 1, du = 4x3 dx will suffice.
CHAPTER 7. TECHNIQUES OF INTEGRATION
480
3x + 4 3 1 1 (b) dx = (2x dx) + 4 dx. Partial fraction decomposition is x2 + 4 2 x2 + 4 x2 + 4 2 unnecessary because the substitution u = x + 4, du = 2x dx will suffice for the first term, while the second term corresponds to formula 24 in Table 7.1.1. 2.1.1 1 x 1 (c) dx = (2x dx). Partial fraction decomposition is unnecessary (x2 + 5)2 2 (x2 + 5)2 because the substitution u = x2 + 5, du = 2x dx will suffice. 1 1 2x3 + 5x dx = (4x3 + 10x dx). Partial fraction decomposition (d) x4 + 5x2 + 6 2 x4 + 5x2 + 6 is unnecessary because the substitution u = x4 + 5x2 + 6, du = 4x3 + 10x dx will suffice. 73. Rewrite the integral as
x5 dx = 10 (x − 1) (x + 1)10
x5 dx = 2 (x − 1)10
u = x2 − 1, x2 = u + 1,
then integrate using
(x2
x4 (x dx) − 1)10
1 u = x dx : 2
2 (u + 1)2 u + 2u + 1 1 1 x4 (x dx) = du = du 2 10 10 (x − 1) 2 u 2 u10 1 1 1 1 (u−8 + 2u−9 + u−10 ) du = − u−7 − u−8 − u−9 + C = 2 14 8 18 1 1 1 = − (x2 − 1)−7 − (x2 − 1)−8 − (x2 − 1)−9 + C 14 8 18
74. The integrand in Problem 53 is an odd function and its definite integral is symmetric about the y-axis. Thus, the definite integral’s value is known to be 0.
7.7
Improper Integrals h
In this exercise set, the symbol “=” is used to denote the fact that L’Hˆ opital’s Rule was applied to obtain the equality.
∞
1. 3
−1
2. −∞
1 dx = lim t→∞ x4
t
x−4 dx = lim
t→∞
3
1 √ dx = lim 3 s→−∞ x
−1 s
x−1/3 dx =
1 − x−3 3
t
= lim
t→∞
3
3 2/3 x s→−∞ 2
1 1 − 81 3t3
−1
lim
= lim
s
s→−∞
=
3 3 2/3 − s 2 2
The integral diverges.
∞
3. 1
1 x0.99
t
dx = lim
t→∞
The integral diverges.
1
x0.01 t→∞ 0.01
x−0.99 dx = lim
t = lim 1
t→∞
1 81
t0.01 1 − 0.01 0.01
CHAPTER 7. TECHNIQUES OF INTEGRATION
496
Using part (d) we see that the fraction of total deaths occurring in the first 34 weeks is
2b
2a b a √ tan−1 √ dt 2 − 2bt + c 2 b 2 t c−b c − b2 0∞ = = tan−1 √ . aπ a π c − b2 √ dt 2 c − b2 −∞ t − 2bt + c With b = 17 and c = 299.13 we find the percentage of total deaths within the first 34 weeks is 17 2 tan−1 √ × 100 = 88.22%. π 299.13 − 172
PROBLEMAS 2.7
7.8 1.
Approximate Integration
Midpoint Rule k 1 2 3 xk 3/2 5/2 7/2 f (xk ) 39/4 95/4 175/4 4 4 − 1 39 95 175 309 2 + + = 77.25 (3x + 2x) dx ≈ = 3 4 4 4 4 1 4 4 (3x2 + 2x) dx = (x3 + x2 ) = 80 − 2 = 78 1
1
2.
Midpoint Rule k 1 2 3 4 xk π/48 π/16 5π/48 7π/48 f (xk ) 0.997859 0.980785 0.94693 0.896873 π/6 π/6 − 0 (0.997859 + 0.980785 + 0.94693 + 0.896873) ≈ 0.500357 cos x dx ≈ 4 0 π/6 π/6 1 1 cos x dx = sin x = −0= 2 2 0 0
3.
Trapezoidal Rule k 0 1 2 3 4 xk 1 3/2 2 5/2 3 f (xk ) 2 35/8 9 133/8 28 3 35 133 3−1 45 ≈ 22.5 (x3 + 1) dx ≈ 2+2 + 2(9) + 2 + 28 = 8 8 8 2 1 3 4 3 x 93 5 3 +x − = 22 (x + 1) dx = = 4 4 4 1 1
7.8. APPROXIMATE INTEGRATION
497
4.
Trapezoidal Rule k 0 1 2 3 4 5 6 xk 0 1/3 2/3 1 4/3 5/3 √ √ √ √ √ √ √ √ √2 f (xk ) 1 2/ 3 5/ 3 2 7/ 3 2 2/ 3 3 2 √ √ √ 2−0 (1 + 2 4/3 + 2 5/3 + 2 2 + 2 7/3 + 2 8/3 + 3) ≈ 2.7954 x + 1 dx ≈ 12 0 2 2 √ 2(x + 1)3/2 2(33/2 ) 2 − ≈ 2.79743 x + 1 dx = = 3 3 3 0 0
5.
Midpoint Rule k 1 2 3 xk 3/2 5/2 7/2 f (xk ) 2/3 2/5 2/7 6 6−1 2 2 1 dx ≈ + 5 3 5 1 x
4 5 9/2 11/2 2/9 2/11 +
2 2 2 + + 7 9 11
=
6086 ≈ 1.75642 3465
Trapezoidal Rule k 0 1 2 3 4 5 xk 1 2 3 4 5 6 f (xk ) 1 1/2 1/3 1/4 1/5 1/6 6 6−1 1 1 1 1 1 1 28 dx ≈ ≈ 1.86667 1+2 +2 +2 +2 + = x 12 2 3 4 5 6 15 1 6.
Midpoint Rule k 1 2 3 4 xk 1/4 3/4 5/4 7/4 f (xk ) 4/7 4/13 4/19 4/25 2 2−0 4 4 4 4 1 27008 dx ≈ + + + ≈ 0.624824 = 3x + 1 4 7 13 19 25 43225 0 Trapezoidal Rule k 0 1 2 3 4 xk 0 1/2 1 3/2 2 f (xk ) 1 2/5 1/4 2/11 1/7 2 2−0 2 1 2 1 1 2161 dx ≈ ≈ 0.701623 1+2 +2 +2 + = 3x + 1 8 5 4 11 7 3080 0
7.
Midpoint Rule k 1 xk 0.05 f (xk ) 1.00125
2 0.15 1.01119
6 7 0.55 0.65 1.14127 1.19269
3 0.25 1.03078 8 0.75 1.25000
4 0.35 1.05948 9 0.85 1.31244
5 0.45 1.09659 10 0.95 1.37931
CHAPTER 7. TECHNIQUES OF INTEGRATION
498
1 0
1−0 (1.00125 + · · · + 1.37931) ≈ 1.1475 x2 + 1 dx ≈ 10
Trapezoidal Rule k 0 1 2 xk 0 0.1 0.2 f (xk ) 1 1.00499 1.0198 6 0.6 1.16619
1 0
8.
7 0.7 1.22066
3 0.3 1.04403 8 0.8 1.28062
4 0.4 1.07703 9 0.9 1.34536
5 0.5 1.11803 10 1.0 1.41421
1−0 [1 + 2(1.00499) + · · · + 2(1.34536) + 1.41421] ≈ 1.14838 x2 + 1 dx ≈ 20
Midpoint Rule k 1 2 3 4 5 xk 1.1 1.3 1.5 1.7 1.9 f (xk ) 0.654981 0.559279 0.478091 0.411241 0.356711 2 1 2−1 √ (0.654981 + · · · + 0.356711) ≈ 0.492061 dx ≈ 3 5 x +1 1 Trapezoidal Rule k 0 1 2 3 4 5 xk 1.0 1.2 1.4 1.6 1.8 2.0 f (xk ) 0.707107 0.605449 0.516811 0.442981 0.382583 0.333333 2 1 2−1 √ [0.707107 + 2(0.605449) + · · · + 2(0.382583) + 0.333333] ≈ 0.493609 dx ≈ 3 10 x +1 1
9.
Midpoint Rule k 1 2 3 4 5 6 xk π/12 π/4 5π/12 7π/12 3π/4 11π/12 f (xk ) 0.0760474 0.180063 0.217033 0.194188 0.128617 0.0429833 π sin x π−0 dx ≈ (0.0760474 + · · · + 0.0429833) ≈ 0.439263 6 0 x+π Trapezoidal Rule k 0 1 2 3 4 5 6 xk 0 π/6 π/3 π/2 2π/3 5π/6 π f (xk ) 0 0.136419 0.206748 0.212207 0.165399 0.0868118 0 π π−0 sin x dx ≈ [1 + 2(0.136419) + · · · + 2(0.0868118) + 0] ≈ 0.42285 12 0 x+π
10.
Midpoint Rule k 1 xk π/24 f (xk ) 0.131652
2 π/8 0.414214
3 5π/24 0.767327
7.8. APPROXIMATE INTEGRATION
π/4 0
tan x dx ≈
π/4 − 0 (0.131652 + 0.414214 + 0.767327) ≈ 0.343793 3
Trapezoidal Rule k 0 1 2 3 xk 0 π/12 π/6 π/4 f (xk ) 0 0.267949 0.57735 1 π/4 π/4 − 0 [0 + 2(0.267949) + 2(0.57735) + 1] ≈ 0.352199 tan x dx ≈ 6 0 11.
Midpoint Rule k 1 2 3 4 5 6 xk 1/6 1/2 5/6 7/6 3/2 11/6 f (xk ) 0.999614 0.968912 0.768409 0.208152 −0.628174 −0.976002 2 2−0 (0.999614 + · · · − 0.976002) ≈ 0.446971 cos x2 dx ≈ 6 0 Trapezoidal Rule k 0 1 2 3 4 5 6 xk 0 1/3 2/3 1 4/3 5/3 2 f (xk ) 1 0.993834 0.90285 0.540302 −0.205507 −0.934546 −0.653644 2 2−0 [1 + 2(0.993834) + · · · + 2(0.934546) − 0.653644] ≈ 0.490037 cos x2 dx ≈ 12 0
12.
Midpoint Rule k 1 2 3 4 5 xk 1/10 3/10 1/2 7/10 9/10 f (xk ) 0.998334 0.985067 0.958851 0.920311 0.870363 1 1−0 sin x dx ≈ (0.998334 + · · · + 0.870363) ≈ 0.946585 x 5 0 Trapezoidal Rule k 0 1 2 3 4 5 xk 0 1/5 2/5 3/5 4/5 1 f (xk ) 1 0.993347 0.973546 0.941071 0.896695 0.841471 1 sin x 1−0 dx ≈ [1 + 2(0.993347) + · · · + 2(0.896695) + 0.841471] ≈ 0.945079 x 10 0
13.
Simpson’s Rule k 0 1 2 3 4 xk 0 √1 √2 √3 4 f (xk ) 1 3 5 7 3 4 √ √ √ √ 4−0 (1 + 4 3 + 2 5 + 4 7 + 3) ≈ 8.6611 2x + 1 dx ≈ 12 0
499
CHAPTER 7. TECHNIQUES OF INTEGRATION
500
4
√
2x + 1 dx
u = 2x + 1, du = 2 dx
0
9
=
u 1
1/2
1 du 2
1 = 2
2 3/2 u 3
9 = 1
26 ≈ 0.86667 3
14.
Simpson’s Rule k 0 1 2 xk 0 π/4 π/2 f (xk ) 0 1/2 1 π/2 1 π/2 − 0 sin2 x dx ≈ 0+4 + 1 ≈ 0.7854 6 2 0 π/2 π/2 1 x sin 2x 1 π 2 (1 − cos 2x) dx = − sin x dx = = ≈ 0.7854 2 2 2 4 0 0 0
15.
Simpson’s Rule k 0 1 2 3 4 xk 1/2 1 3/2 2 5/2 f (xk ) 2 1 2/3 1/2 2/5 5/2 5/2 − 1/2 2 1 1 2 dx ≈ 2 + 4(1) + 2 +4 + ≈ 1.6222 12 3 2 5 1/2 x
16.
Simpson’s Rule k 0 1 2 3 4 5 6 xk 0 5/6 5/3 5/2 10/3 25/6 5 f (xk ) 1/2 6/17 3/11 2/9 3/16 6/37 1/7 5 6 1 5−0 1 3 2 3 6 1 dx ≈ +4 +2 +4 +2 +4 + ≈ 1.2535 18 2 17 11 9 16 37 7 0 x+2
17.
Simpson’s Rule k 0 1 2 3 4 xk 0 1/4 1/2 3/4 1 f (xk ) 1 16/17 4/5 16/25 1/2 1 1 16 4 16 1−0 1 dx ≈ 1 + 4 + 2 + 4 + ≈ 0.7854 2 12 17 5 25 2 0 1+x
18.
Simpson’s Rule k 0 1 2 xk −1 0 √1 √ f (xk ) 2 1 2 1 √ 1+1 √ [ 2 + 4(1) + 2] ≈ 2.2761 x2 + 1 dx ≈ 6 −1
7.8. APPROXIMATE INTEGRATION 19.
501
Simpson’s Rule k 0 1 2 3 4 5 6 xk 0 π/6 π/3 π/2 2π/3 5π/6 π √ √ f (xk ) 0 3/7π 3 8/8π 2/3π 3 3/10π 3/11π 0 √ √ π 3 2 3 π−0 sin x 3 3 3 3 dx ≈ 0+4 +4 +4 +2 +2 +0 18 7π 8π 3π 10π 11π 0 x+π ≈ 0.4339
20.
Simpson’s Rule k 0 1 2 3 4 xk 0 1/4 1/2 3/4 1 f (xk ) 1 0.8776 0.7602 0.6479 0.5403 1 √ 1−0 [1 + 4(0.8776) + 2(0.7602) + 4(0.6479) + 0.5403] ≈ 0.7635 cos x dx ≈ 12 0
21.
Simpson’s Rule k 0 1 2 3 4 xk 2 2.5 3 3.5 4 f (xk ) 3.1623 4.2573 5.4772 6.8099 8.2462 4 4−2 [3.1623 + 4(4.2573) + 2(5.4772) + 4(6.8099) + 8.2462] ≈ 11.1053 x3 + x dx ≈ 12 2
22.
Simpson’s Rule k 0 1 2 xk π/4 3π/8 π/2 f (xk ) 0.3694 0.3420 0.3333 π/2 1 π/2 − π/4 dx ≈ [0.3694 + 4(0.3420) + 0.3333] ≈ 0.2711 2 + sin x 6 π/4 1 ; x+3
2 . Since f (x) decreases on [−1, 2], (x + 3)3 (1/4)(2 + 1)3 f (x) ≤ f (−1) = 1/4 on the interval. Taking M = 1/4, we want < 0.005 or 24n2 225 = 56.25. Take n = 8 to obtain the desired accuracy. n2 > 4
23. f (x) =
f (x) = −
1 ; (x + 3)2
f (x) =
24. f (x) = sin2 x;
f (x) = 2 sin x cos x = sin 2x; f (x) = 2 cos 2x. Since |f (x)| = 2| cos 2x| ≤ 2(1.5 − 0)3 < 0.0001 or n2 > 5625 = 752 . Take 2 for all x, we take M = 2. Then we want 12n2 n = 76 to obtain the desired accuracy. 6x2 − 2 . To determine an upper bound (1 + x2 )3 24x(1 − x2 ) . Setting f (x) = 0 we obtain the for |f (x)| on [0, 2], we compute f (x) = (1 + x2 )4
25. f (x) =
1 ; 1 + x2
f (x) = −
2x ; (1 + x2 )2
f (x) =
CHAPTER 7. TECHNIQUES OF INTEGRATION
502
1 22 , and f (2) = , 2 125 3 2(2 − 0) we see that |f (x)| ≤ 2 on the interval. Taking M = 2 we want < 0.005 or 12n2 2 1 800 ≈ 267. For n = 17, the Trapezoidal Rule gives n2 > dx ≈ 1.11 to two decimal 2 3 0 1+x places. critical numbers 0 and 1 on [0, 2]. Comparing f (0) = −2, f (1) =
26.
Trapezoidal Rule k 0 1 xk −2 −1 f (xk ) 0.01 0.1
2 −2
10x dx ≈
3 4 1 2 10 100
2+2 [0.01 + 2(0.1) + 2(1) + 2(10) + 100] ≈ 61.105 8
24 . Since f (4) (x) is decreasing on x5 24(3 − 1)5 [1, 3], |f (4) (x)| ≤ f (4) (1) = 24 for all x in [1, 3]. Taking M = 24, we want < 10−5 180n4 1, 280, 000 ≈ 426, 667. To have the desired accuracy with Simpson’s Rule, we need or n4 > 3 2 n ≥ 26. To obtain the required n for the Trapezoidal Rule, we note that f (x) = 3 x 2(3 − 1)3 is decreasing on [1, 3]. We thus take M = f (1) = 2. We want < 10−5 or 12n2 400, 000 ≈ 133, 333. For the Trapezoidal Rule to have the desired accuracy, we need n2 > 3 n ≥ 366.
27. f (x) = −
1 ; x2
2 0 1
f (x) =
2 ; x3
f (x) = −
6 ; x4
f (4) (x) =
2 8 48 1 ; f (x) = − ; f (x) = ; f (x) = − ; 2 3 2x + 1 (2x + 1) (2x + 1) (2x + 1)4 384 f (4) (x) = . Since f (4) (x) is decreasing on [0, 3], |f (4) (x)| ≤ f (4) (0) = 384 for (2x + 1)5 2 384(3 − 0)5 all x in [0, 3]. Taking M = 384, we have E6 ≤ = = 0.4. 4 180(6 ) 5
28. f (x) =
29. Since n = 5 is odd we cannot use Simpson’s Rule. Because the Midpoint Rule does not readily work with tabular data we use the Trapezoidal Rule:
2.30 2.05
f (x) dx ≈
2.30 − 2.05 [4.91 + 2(4.80) + 2(4.66) + 2(4.41) + 2(3.93) + 3.58] = 1.10225. 10
30. Since the subintervals are of unequal widths, Simpson’s Rule as derived in the text cannot be
7.8. APPROXIMATE INTEGRATION
503
applied. We use trapezoids to approximate the integral.
1.20 0
4
31.
(2x + 5) dx = (x2 + 5x)
0
4
n=2:
0
4
n=4: 0
4
32.
−0.55 − 0.16 −0.72 − 0.55 + (0.2 − 0.1) 2 2 0.62 + 0.78 −0.16 + 0.62 + (0.6 − 0.4) + (0.4 − 0.2) 2 2 0.78 + 1.34 1.34 + 1.47 + (0.8 − 0.6) + (0.9 − 0.8) 2 2 1.47 + 1.61 1.61 + 1.51 + (1.0 − 0.9) + (1.2 − 1.0) 2 2 = 0.9055
f (x) dx ≈ (0.1 − 0)
(2x + 5) dx ≈
4
n=2:
0
n=4: 0
4
0
= 36
4−0 [(2 · 1 + 5) + (2 · 3 + 5)] = 36 2
4−0 [(2 · 0.5 + 5) + (2 · 1.5 + 5) + (2 · 2.5 + 5) + (2 · 3.5 + 5)] 4 = 36
(2x + 5) dx ≈
(2x + 5) dx = (x2 + 5x)
0
4
(2x + 5) dx ≈ (2x + 5) dx ≈
1
33. (a) I = −1
4 0
= 36
4−0 [(2 · 0 + 5) + 2(2 · 2 + 5) + (2 · 4 + 5)] = 36 4 4−0 [(2 · 0 + 5) + 2(2 · 1 + 5) + 2(2 · 2 + 5) + 2(2 · 3 + 5) 8 + (2 · 4 + 5)] = 36
(x3 + x2 ) dx =
1 4 1 3 x + x 4 3
1 = −1
2 3
1 − (−1) [f (−7/8) + f (−5/8) + · · · + f (5/8) + f (7/8)] 8 75 325 735 1 49 21 + + ··· + + = = 4 512 512 512 512 32
(b) M8 =
1 − (−1) [f (−1) + 2f (−3/4) + · · · + 2f (3/4) + f (1)] 16 1 9 63 11 = 0+2 + ··· + 2 +2 = 8 64 64 16
2
2 1 1 ; E8 =
− T8
= (d) E8 =
− M8
= 3 96 3 48 (c) T8 =
The error for the Midpoint Rule is one half the error for the Trapezoidal Rule.
CHAPTER 7. TECHNIQUES OF INTEGRATION
504
3
34. −1
(x3 − x2 ) dx =
3
−1 3
n=2: n=4: −1
1 4 1 3 x − x 4 3
3 = −1
32 3
(x3 − x2 ) dx ≈
3+1 32 {[(−1)3 − (−1)2 ] + 4(13 − 12 ) + (33 − 32 )} = 6 3
(x3 − x2 ) dx ≈
3+1 {[(−1)3 − (−1)2 ] + 4(03 − 02 ) + 2(13 − 12 ) 12 32 + 4(23 − 22 ) + (33 − 32 )} = 3
35. The exact value is
b
(c1 x + c0 ) dx = a
c
1
x2 + c 0 x
b
=
c1 2 (b − a2 ) + c0 (b − a) 2
2 a b−a = [(c1 a + c0 ) + (c1 b + c0 )]. 2
The Trapezoidal Rule gives b−a b−a 2(b − a) (c1 a + c0 ) + 2 c1 a + + c0 + 2 c1 a + + c0 + · · · 2n n n n(b − a) (n − 1)(b − a) + 2 c1 a + + c0 + c1 a + + c0 n n b−a b−a (1 + 2 + · · · + n − 1) + c1 (b − a) = 2nc1 a + 2nc0 + 2c1 2n n b−a b−a (n − 1)n = 2nc1 a + 2nc0 + 2c1 + c1 (b − a) 2n n 2 b−a [2nc1 a + 2nc0 + c1 (b − a)(n − 1) + c1 (b − a)] = 2n b−a [2nc1 a + 2nc0 + nc1 b − nc1 a − c1 b + c1 a + c1 b − c1 a] = 2n b−a (c1 a + c0 + c1 b + c0 ). = 2n Since the graph of f (x) = c1 x + c0 is a straight line and the Trapezoidal Rule uses straight line approximations to the curve, it will give the exact value. 36. From the derivation of Simpson’s Rule in the text, it suffices to show that it will give the h (c3 x3 + c2 x2 + c1 x + c0 ) dx for n = 2. In this case, the exact value is exact value of −h
c 2c c3 4 c2 3 c1 2 c2 c1 3 2 3 h + h + h + c0 h − (−h)4 + (−h)3 + (−h)2 + c0 (−h) = h + 2c0 h, 4 3 2 4 3 2 3
7.8. APPROXIMATE INTEGRATION
505
and Simpson’s Rule gives h+h [c3 (−h)3 + c2 (−h)2 + c1 (−h) + c0 ] + 4(c0 ) + (c3 h3 + c2 h2 + c1 h + c0 ) 6 2c2 3 h h + 2c0 h. = (2c2 h2 + 6c0 ) = 3 3
4
37. 1
f (x) dx ≈
4−1 [1.3 + 4(1.5) + 2(3) + 4(3.3) + 2(2.2) + 4(2.4) + 1.9] ≈ 7.06667 3(6)
9−0 67 [0 + 2(3) + 2(1) + 2(3) + 2(5) + 2(3) + 2(7) + 2(4) + 2(6) + 3] = 18 2 The Trapezoidal Rule gives the exact value of the area in this case because the curve segments are straight lines.
38. A =
18.6 − 0 [0 − 4(5.8) + 2(7.3) + 4(6.9) + 2(8.7) + 4(8.8) + 2(10.3) + 4(14.5) 3(10) + 2(15) + 4(10.4) + 0] = 166.284
39. A ≈
The volume of the pond is 4(166.284) = 665.136 ft3 and the number of gallons of water is 7.48(665.136) = 4975.22 gallons. 3(570π) 3(570) 4.5 − 1 + 40. I ≈ [1(0.3) + 2(1.5)2 (0.5) + 2(2)2 (0.62) + 2(2.5)2 (0.7) 2(32) 32 14 + 2(3)2 (0.6) + 2(3.5)2 (0.5) + 2(4)2 (0.27) + (4.5)2 (0)] ≈ 83.94 + 53.44(11.99) ≈ 724.69 41.
Simpson’s Rule k 0 1 xk −5 −4 f (xk ) 0 3.55936
5
2 −3 4.38712
3 −2 4.79112
4 −1 4.96403
5 0 5 -5
6 7 1 2 4.96403 4.79112
8 3 4.38712
9 4 3.55936
5
10 5 0
5 − (−5) 5 [0 + 4(3.55936) + 2(4.38712) + · · · (52.5 − |x|2.5 )2 dx ≈ 30 −5 5
+ 2(4.38712) + 4(3.55936) + 0] ≈ 41.4028 42. From the graph we define f (0) = 2. 2
Simpson’s Rule k 0 1 xk 0 π/5 f (xk ) 2 1.71614
2 2π/5 1.93889
3 3π/5 1.90975
4 5 4π/5 π 1.26302 1
2
CHAPTER 7. TECHNIQUES OF INTEGRATION
506 6 6π/5 1.13489
2π 0
7 7π/5 1.80195
8 8π/5 1.77706
9 9π/5 1.04954
10 2π 1
2π − 0 [2 + 4(1.71614) + 2(1.93889) + · · · 30 + 2(1.77706) + 4(1.04954) + 1] ≈ 9.45351 k 1/x 1 e 1 dx = lim dx x = , dx = − 2 dt 5/2 k→∞ 1 x t t 0 1/k 1 t e 1 1/2 t = lim (−t e dt) = t1/2 et dt − 2 dt = k→∞ 1 t (1/t)5/2 1 0 1 dx = t1/2 et dt 0 √ 1 1/4 1 1/2 3 3/4 1−0 0+4 e e e +4 +e +2 ≈ 12 2 2 2
(1 + | sin x|x ) dx ≈
∞
e1/x x5/2
∞
e1/x x5/2
43. (a) 1
(b) 1
1 (2.5681 + 2.3316 + 7.3335 + 2.7183) ≈ 1.2460 12 2 √ √ √ √ 2−0 (1 + 4 1.0625 + 2 2 + 4 6.0625 + 17) ≈ 3.6539 s= 1 + x4 dx ≈ 12 0 ≈
44. y = x2 ;
45. y = 2x 1 1 − 0 1 + 4x2 dx ≈ s= [1 + 4(0)2 ]1/2 + 2[1 + 4(0.1)2 ]1/2 + 2[1 + 4(0.2)2 ]1/2 + · · · 20 0 + 2[1 + 4(0.9)2 ]1/2 + [1 + 4(1)2 ]1/2 √ √ √ √ 1 (1 + 2 1.04 + 2 1.16 + · · · + 2 4.24 + 5) ≈ 1.4804 = 20 2 1 46. y = ; s = 1 + 1/x2 dx. Using Simpson’s Rule with n = 6 we obtain x 1 1 √ S6 = 1 + 1 + 4 1 + 36/49 + 2 1 + 9/16 + 4 1 + 4/9 + 2 1 + 9/25 18 + 4 1 + 36/121 + 1 + 1/4 4√ 1√ 4√ 2√ 4√ 1√ 1 √ 2+ 85 + 25 + 13 + 34 + 157 + 5 ≈ 1.2220. = 18 7 2 3 5 11 2
47. y = x;
2
S = 2π 0
1 2 x 1 + x2 dx = π 2
2 0
x2
1 + x2 dx. Using the Midpoint Rule with
n = 5, we obtain √ 9 2 1 1 9 49 49 81 81 + + 2+ + ≈ 4.76741. M5 = 1+ 1+ 1+ 1+ 5 25 25 25 25 25 25 25 25
CHAPTER 7 IN REVIEW
507
Then S ≈ πM5 ≈ 14.9772. 1 dx = 2y; S = 2π 48. (y 2 + 1) 1 + 4y 2 dy. Using Simpson’s Rule with n = 6, we obtain dy −1 S6 =
1 − (−1) √ [2 5 + 4(4/9 + 1) 1 + 16/9 + 2(1/9 + 1) 1 + 4/9 3.6 √ √ + 4(1) 1 + 2(1/9 + 1) 1 + 4/9 + 4(4/9 + 1) 1 + 16/9 + 2 5] ≈ 4.17168.
Then S ≈ 2πS6 ≈ 26.2114. 49. (a) Approximating the graph on each interval between integers with line segments and using the Pythagorean theorem, we have s < 12 + 22 + 12 + 42 + 12 + 32 + 12 + 12 + 12 + 32 + 12 + 12 + 12 + 22 √ √ √ √ √ √ √ = 5 + 17 + 10 + 2 + 10 + 2 + 5 ≈ 17.75. (b) Since y (xi ) = 0 for xi = 1, 2, . . . , 8, the formula for arc length gives 8 dx = 7 which is simply the length of the interval. 1
Li(x) = lim 50. (a) lim x→∞ x/ ln x x→∞
x
ln x 2
x x
x+ = lim
x→∞
1 dt ln t
2
1 dt ln t
x
1 ln x(1/ ln x) + x h = lim x→∞ 1 h
x 2
8
1 + (y )2 dx =
1
1 dt ln t
1 + 1/ ln x =1 x→∞ 1
= lim
100
1 dt ≈ 29.09. ln t 2 When x = 100, x/ ln x ≈ 21.71. The number of primes less than 100 is 25.
(b) Implementing Simpson’s Rule with n = 98 on a computer, we obtain
PROBLEMAS DEReview REPASO DE LA UNIDAD 2 Chapter 7 in A. True/False 1. True 2. False; use u = a tan θ. 3. True 4. True 5. True 6. False;
A B x2 + =1+ . (x + 1)2 x + 1 (x + 1)2
CHAPTER 7. TECHNIQUES OF INTEGRATION
508 7. False;
1 B D A C + + = + . (x2 − 1)2 x − 1 (x − 1)2 x + 1 (x + 1)2
8. False; it is used n times. 9. False; substituting u = 9 − x2 will work. 10. True 11. True 12. False; let f (x) = x and g(x) = −x. 13. True
∞ −∞
s→−∞
a
f (x) dx = lim
14. False;
t
f (x) dx + lim s
t→∞
f (x) dx. a
15. False; the infinite discontinuity occurs at x = 1/e ≈ 0.3679, which is outside of [1/2, 1].
6.
sin x ln(sin x) dx
u = ln(sin x), du = cot x dx; dv = sin x dx, v = − cos x cos2 x dx = − cos x ln(sin x) + cos x cot x dx = − cos x ln(sin x) + sin x 1 − sin2 x = − cos x ln(sin x) + dx sin x = − cos x ln(sin x) + csc x dx + (− sin x) dx = cos x − cos x ln(sin x) + ln | csc x − cot x| + C
B. C. Exercises
√
1.
2.
e
√
3.
1 dx x+9
√
x+1
dx
x x2 + 4
√
x + 9, x = (u − 9)2 , dx = 2(u − 9) du 1 2(u − 9) du = 2 du − 18 du = 2u = 18 ln |u| + C = u u √ √ √ √ = 2 x + 18 − 18 ln( x + 9) + C = 2 x − 18 ln( x + 9) + C1 u=
√
x + 1, x = t2 − 1, dx = 2t dt = et (2t dt) = 2tet dt u = 2t, du = 2 dt; dv = et dt, v = et √ √ √ t = 2te − 2et dt = 2tet − 2et + C = 2 x + 1 e x+1 − 2e x+1 + C t=
dx =
1 2
(x2 + 4)−1/2 (2x dx) =
1 (x2 + 4)1/2 + C = x2 + 4 + C 2 1/2
CHAPTER 7. TECHNIQUES OF INTEGRATION
510 √
4.
5.
1 x2
+4
dx
2 sec2 θ √ x = 2 tan θ, dx = 2 sec2 θ dθ = dθ 4 tan2 θ + 4 sec2 θ dθ = sec θ dθ = ln | sec θ + tan θ| + C = sec θ
√
x2 + 4 x
+ + C = ln x2 + 4 + x + C1 = ln
2 2
1 dx (x2 + 4)3
2
x = 2 tan θ, dx = 2 sec θ dθ 1 1 sec2 θ dθ = = cos4 θ dθ 32 sec6 θ 32
=
2 sec2 θ dθ (4 tan2 θ + 4)3
4 + x2
x
θ
2
4 + x2
x
θ
2
See Section 7.4, 2.4, Example 5 1 1 3 θ+ sin 2θ + sin 4θ + C 256 128 1024 3 1 1 x = tan−1 + sin θ cos θ + sin θ cos θ(cos2 θ − sin2 θ) + C 256 2 64 256 x x 4 − x2 x 3 tan−1 + + = +C 256 2 32(x2 + 4) 128(x2 + 4) x2 + 4 x x x x3 3 tan−1 + + − +C = 2 2 2 256 2 32(x + 4) 32(x + 4) 128(x2 + 4)2
=
x2 + 4 − 4 1 x dx = dx − 4 dx = x − 2 tan−1 + C 2 2 x +4 x +4 2 2 x +4 1 4 7. dx = dx + 4 dx = x − + C 2 2 x x x
6.
x2 dx = 2 x +4
8. Write
3x − 1 A B C = + + . Then 3x − 1 = A(x2 − 4) + B(x2 + 2x) + C(x2 − 2x). x(x2 − 4) x x−2 x+2
Setting x = 0, x = 2, and x = −2 gives A = 1/4, B = 5/8, and C = −7/8. Thus 1 1 5 1 7 1 3x − 1 dx = dx + dx − dx 2 x(x − 4) 4 x 8 x−2 8 x+2 1 5 7 = ln |x| + ln |x − 2| − ln |x + 2| + C. 4 8 8
x−5 1 dx = x2 + 4 2 √ 3 x + 27 10. dx x 9.
2x dx − 5 x2 + 4 √ 3
x 1 1 5 dx = ln(x2 + 4) − tan−1 + C x2 + 4 2 2 2
x + 27, x = u3 − 27, dx = 3u2 du 3 1 u u − 27 + 27 2 (3u du = 3 du + 81 du = du) = 3 3 3 3 u − 27 u − 27 u − 27 u=
CHAPTER 7 IN REVIEW Write
511
1 A Bu + C = + . Then u3 − 27 u − 3 u2 + 3u + 9
1 = A(u2 + 3u + 9) + (Bu + C)(u − 3) = (A + B)u2 + (3A − 3B + C)u + (9A − 3C). Solving
3A − 3B + C = 0
A+B =0
9A − 3C = 1
gives A = 1/27, B = −1/27, and C = −2/9. Thus √ 3 1 1 1 u 2 1 x + 27 dx = 3 du + 81 du − du − du x 27 u − 3 27 u2 + 3u + 9 9 u2 + 3u + 9 3 1 2u + 3 − 3 = 3u + 3 ln |u − 3| − du − 18 du 2 2 2 u + 3u + 9 u + 3u + 9 3 2u + 3 27 1 = 3u + 3 ln |u − 3| − du − du 2 u2 + 3u + 9 2 (u + 3/2)2 + 27/4 √ 3 2u + 3 = 3u + 3 ln |u − 3| − ln(u2 + 3u + 9) − 3 3 tan−1 √ + C 2 3 3 √ √ 3 = 3 3 x + 27 + 3 ln | 3 x + 27 − 3| − ln[(x + 27)2/3 + 3(x + 27)1/3 + 9] 2 √ 3 √ 2 x + 27 + 3 √ − 3 3 tan−1 + C. 3 3 11. 12.
(ln x)9 dx x
u = ln x, du =
(ln 3x)2 dx
u = (ln 3x)2 , du =
1 dx x
=
2 ln 3x dx; x
u9 du =
1 10 1 u +C = (ln x)10 + C 10 10
dv = dx, v = x
1 u = ln 3x, du = dx; dv = dx, v = x x = x(ln 3x)2 − 2 x ln 3x − dx = x(ln 3x)2 − 3x ln 3x + 2x + C 2
= x(ln 3x) − 2
13.
t sin−1 t dt
ln 3x dx
u = sin−1 t, du = √ = = = = =
1 dt; 1 − t2
2
dv = t dt, v =
1 2 t 2
1
θ t dt t = sin θ, dt = cos θ dθ 1 − t2 2 1−t sin2 θ 1 1 2 −1 1 1 t sin t − cos θ dθ = t2 sin−1 t − sin2 θ dθ 2 2 cos θ 2 2 1 2 −1 1 2 −1 1 1 1 t sin t − (1 − cos 2θ) dθ = t sin t − θ − sin 2θ + C 2 4 2 4 2 1 2 −1 1 1 t sin t − sin−1 t + sin θ cos θ + C 2 4 4 1 2 −1 1 1 t sin t − sin−1 t + t 1 − t2 + C 2 4 4
1 2 −1 1 t sin t − 2 2
√
t
CHAPTER 7. TECHNIQUES OF INTEGRATION
512 14.
ln x dx (x − 1)2
u = ln x, du = =−
ln x + x−1
1 dx; x
dv =
1 1 dx, v = − (x − 1)2 x−1
1 dx x(x − 1)
A B 1 = + . Then 1 = A(x − 1) + Bx. Setting x = 0 and x = 1 gives x(x − 1) x x−1 A = −1 and B = 1. Thus 1 1 ln x ln x − dx + dx dx = − 2 (x − 1) x−1 x x−1
x − 1
ln x ln x
+ C. =− − ln |x| + ln |x − 1| + C = − + ln
x−1 x−1 x
Write
15.
(x + 1)3 (x − 2) dx
= =
16. Write
u = x + 1, x = u − 1, dx = du 3 u (u − 3) du = (u4 − 3u3 ) du
1 3 1 5 3 4 u − u + C = (x + 1)5 − (x + 1)4 + C 5 4 5 4
1 A B C D = + . + + 3 2 3 (x + 1) (x − 2) x + 1 (x + 1) (x + 1) x−2
Then 1 = A(x + 1)2 (x − 2) + B(x + 1)(x − 2) + C(x − 2) + D(x + 1)3 = (A + D)x3 + (B + 3D)x2 + (−3A − B + C + 3D)x + (−2A − 2B − 2C + D). Solving
A+D =0 −3A − B + C + 3D = 0
B + 3D = 0 −2A − 2B − 2C + D = 1
gives A = −1/27, B = −1/9, C = −1/3, and D = 1/27. Thus 1 1 1 1 1 1 1 1 1 dx = − dx − dx dx − dx + (x + 1)3 (x − 2) 27 x + 1 9 (x + 1)2 3 (x + 1)3 27 x − 2 1 1 1 1 1 1 = − ln |x + 1| + ln |x − 2| + C + + 27 9 x+1 6 (x + 1)2 27
x − 2
1 1 1
+ ln
+ = + C.
27 x+1 9(x + 1) 6(x + 1)2 17.
ln(x2 + 4) dx
u = ln(x2 + 4), du =
dv = dx, v = x
2 x +4−4 2x2 2 dx = x ln(x dx + 4) − 2 x2 + 4 x2 + 4 x 1 = x ln(x2 + 4) − 2 dx + 8 dx = x ln(x2 + 4) − 2x + 4 tan−1 + C 2 x +4 2
= x ln(x2 + 4) −
2x dx; +4
x2
CHAPTER 7 IN REVIEW 18.
8te2t
2
u = 2t2 , du = 4t dt =
19. Write
x4
513
2
2eu du = 2eu + C = 2e2t + C
B D 1 A C + = + 2+ . 3 2 + 10x + 25x x x x + 5 (x + 5)2
Then 1 = Ax(x + 5)2 + B(x + 5)2 + Cx2 (x + 5) + Dx2 = (A + C)x3 + (10A + B + 5C + D)x2 + (25A + 10B)x + 25B. Solving
A+C =0 25A + 10B = 0
10A + B + 5C + D = 0 25B = 1
gives A = −2/125, B = 1/25, C = 2/125, and D = 1/25. Thus 1 1 1 1 1 1 2 2 1 dx + dx + dx = − dx + dx x4 + 10x3 + 25x2 125 x 25 x2 125 x + 5 25 (x + 5)2 1 1 1 1 2 2 ln |x| − ln |x + 5| − =− + +C 125 25 x 125 25 x + 5 20.
1 dx = x2 + 8x + 25
21. Write
1 1 x+4 dx = tan−1 +C (x + 4)2 + 9 3 3
A B x C = + . + x3 + 3x2 − 9x − 27 x + 3 (x + 3)2 x−3
Then x = A(x2 − 9) + B(x − 3) + C(x + 3)2 = (A + C)x2 + (B + 6C)x + (−9A − 3B + 9C). Solving
A+C =0
B + 6C = 1
− 9A − 3B + 9C = 0
gives A = −1/12, B = 1/2, and C = 1/12. Thus 1 1 1 1 1 1 x dx = − dx + dx dx + 3 2 2 x + 3x − 9x − 27 12 x+3 2 (x + 3) 12 x−3 1 1 1 1 = − ln |x + 3| − ln |x − 3| + C + 12 2 x+3 12
x − 3
1 1
− ln
+ C. =
12 x+3 2(x + 3) 22. Write
x+1 A B Cx + D = + + 2 . (x2 − x)(x2 + 3) x x−1 x +3
Then x + 1 = A(x − 1)(x2 + 3) + Bx(x2 + 3) + (Cx + D)(x2 − x) = (A + B + C)x3 + (−A − C + D)x2 + (3A + 3B − D)x − 3A. Solving
A+B+C =0
−A − C + D = 0
3A + 3B − D = 1
−3A = 1
CHAPTER 7. TECHNIQUES OF INTEGRATION
514
gives A = −1/3, B = 1/2, C = −1/6, and D = −1/2. Thus
23. 24.
25.
1 1 1 1 1 x 1 1 x+1 dx = − dx + dx − dx − dx (x2 − x)(x2 + 3) 3 x 2 x−1 6 x2 + 3 2 x2 + 3 1 1 1 x 1 = − ln |x| + ln |x − 1| − ln(x2 + 3) − √ tan−1 √ + C. 3 2 12 2 3 3
sin2 t dt = cos2 t
tan2 t dt =
sin3 θ dθ = (cos θ)3/2
(sec2 t − 1) dt = tan t − t + C
1 − cos2 θ sin θ dθ u = cos θ, du = − sin θ dθ (cos θ)3/2 u−1/2 1 − u2 u3/2 + +C =− du = (u1/2 − u−3/2 ) du = 3/2 3/2 1/2 u 2 2 = (cos θ)3/2 + +C 3 (cos θ)1/2
tan10 x sec4 x dx =
tan10 x(tan2 x + 1) sec2 x dx u = tan x, du = sec2 x dx 1 13 1 u + u11 + C = u10 (u2 + 1) du = (u12 + u10 ) du = 13 11 1 1 13 11 tan x + tan x + C = 13 11
26. Write
x sin x x tan x = . Now cos x cos2 x
cos2 x + 2x cos x sin x x , du = dx; dv = sin x dx, v = − cos x cos2 x cos4 x cos2 x + 2x cos x sin x x x sin x x + dx = − + sec x dx + 2 dx =− cos x cos3 x cos x cos2 x x x sin x =− + ln | sec x + tan x| + 2 dx cos x cos2 x x sin x x Solving for the integral, dx = − ln | sec x + tan x| + C. 2 cos x cos x x sin x dx cos2 x
u=
27.
y cos y dy
28.
u = y, du = dy; dv = cos y dy, v = sin y = y sin y − sin y dy = y sin y + cos y + C
1 x2 sin x3 dx = − cos x3 + C 3
CHAPTER 7 IN REVIEW
2
29.
515
3
2
2
(1 + sin t)(1 − sin t) cos t dt =
(1 + sin t) cos t dt =
(1 − sin4 t) cos t dt
u = sin t, du = cos t dt
1 1 (1 − u4 ) du = u − u5 + C = sin t − sin5 t + C 5 5 sec2 θ sec2 θ sec3 θ sec θ tan θ dθ = dθ = sec θ tan θ dθ 30. 2 tan θ sec2 θ − 1 tan θ =
u = sec θ, du = sec θ tan θ dθ 2 u −1+1 1 u2 du = du = du + du = 2 2 2 u −1 u −1 u −1 A B 1 = + . Then 1 = A(u + 1) + B(u − 1). Setting u = 1 and u = −1 Write 2 u −1 u−1 u+1 gives A = 1/2 and B = −1/2. Thus 1 1 1 1 1 1 sec3 θ dθ = du + du − du = u + ln |u − 1| − ln |u + 1| + C tan θ 2 u−1 2 u+1 2 2
1
sec θ − 1
1
u − 1
+ C = sec θ + ln
+ C. = u + ln
2 u + 1
2 sec θ + 1
ew (1 + ew )5 dw
31.
u = 1 + ew , du = ew dw
u5 du =
=
(x − 1)e−x dx
32.
cot 4x dx =
33.
u = x − 1, du = dx; dv = e−x dx, v = −e−x = −(x − 1)e−x + e−x dx = −(x − 1)e−x − e−x + C
3
1 6 1 u + C = (1 + ew )6 + C 6 6
2
(csc 4x − 1) cot 4x dx =
csc 4x csc 4x cot 4x dx −
cot 4x dx
u = csc 4x, du = −4 csc 4x cot 4x dx 1 1 1 1 u du − ln | sin 4x| + C = − u2 − ln | sin 4x| + C =− 4 4 8 4 1 1 = − csc2 4x − ln | sin 4x| + C 8 4 34. (3 − sec x)2 dx = (9 − 6 sec x + sec2 x) dx = 9 − 6 ln | sec x + tan x| + tan x + C
π/4
35. 0
π/4
1 cos x tan x dx = cos x sin x dx = 2 0 1 1 = − (0 − 1) = 4 4 2
π/4 0
1 sin 2x dx = − cos 2x 4
π/4 0
CHAPTER 7. TECHNIQUES OF INTEGRATION
516
π/3
36. 0
37.
π/3
sin x dx u = cos x, du = − sin x dx cos x 0 1/2 1/2 (1 − u2 )2 1 3 du = − − 2u + u du =− u u 1 1 1/2 1 1 15 3 33 = − ln − = − ln |u| − u2 + u4 − − = ln 2 − 4 2 64 4 64 1
sin4 x tan x dx =
(1 − cos2 x)2
(sin x)(1 − sin x) sin x − sin2 x dx dx = (1 + sin x)(1 − sin x) 1 − sin2 x sin x sin2 x 1 sin x 2 = − dx = − tan x dx cos2 x cos2 x cos x cos x = [sec x tan x − (sec2 x − 1)] dx = sec x − tan x + x + C
sin x dx = 1 + sin x
Alternatively, the substitution u = tan
38.
cos x dx 1 + sin x
sin x 2 dx = x + + C. 1 + sin x 1 + tan x/2
u = 1 + sin x, du = cos x dx =
39. Write
x leads to the equivalent solution 2
1 du = ln |u| + C = ln(1 + sin x) + C u
1 A B C = + + . (x + 1)(x + 2)(x + 3) x+1 x+2 x+3
Then 1 = A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2). Setting x = −1, x = −2, and x = −3 gives A = 1/2, B = −1, and C = 1/2. Thus
1 0
1 1 1 1 1 1 dx − dx + dx 2 0 x+3 0 x+1 0 x+2 1 1 1 1 1 = ln |x + 1| − ln |x + 2| + ln |x + 3| 2 2 0 0 0 1 5 3 1 = (ln 2 − ln 1) − (ln 3 − ln 2) + (ln 4 − ln 3) = ln 2 − ln 3. 2 2 2 2
1 1 dx = (x + 1)(x + 2)(x + 3) 2
1
CHAPTER 7 IN REVIEW
ln 2
40.
√
ex + 1 dx
ln 3
ex cos 3x dx
41.
517 √
2u du u2 − 1 √3 2 √3 √3 √3 2u2 u −1+1 1 du = 2 du = 2 du = du + 2 2−1 2−1 2−1 u u u 2 2 2 2 √ √ 3 3 √ 3 1/2 −1/2 = 2u du + 2 du +2 u−1 u+1 2 2 2 √3 √ 3 √ = 2 3 − 4 + ln |u − 1| − ln |u + 1| 2 2 √ √ √ √ √ 3−1 + ln 3 = 2 3 − 4 + ln( 3 − 1) − [ln( 3 + 1) − ln 3] = 2 3 − 4 + ln √ 3+1 u=
ex + 1, x = ln(u2 − 1), dx =
1 u = ex , du = ex dx; dv = cos 3x dx, v = sin 3x 3 1 x 1 ex sin 3x dx = e sin 3x − 3 3
1 u = ex , du = ex dx; dv = sin 3x dx, v = − cos 3x 3 1 1 1 1 ex cos 3x dx = ex sin 3x − − ex cos 3x + 3 3 3 3 3 x 1 e sin 3x + ex cos 3x + C. Solving for the integral, ex cos 3x dx = 10 10 42.
x(x − 5)9 dx
43.
cos(ln t) dt
u = x − 5, du = dx 1 11 1 10 u + u +C = (u + 5)u9 du = (u10 + 5u9 ) du = 11 2 1 1 (x − 5)11 + (x − 5)10 + C = 11 2
sin(ln t) dt; u = cos(ln t), du = − t = t cos(ln t) + sin(ln t) dt
dv = dt, v = t
cos(ln t) dt; dv = dt, v = t t = t cos(ln t) + t sin(ln t) − cos(ln t) dt u = sin(ln t), du =
Solving for the integral,
cos(ln t) dt =
1 1 t cos(ln t) + t sin(ln t) + C. 2 2
CHAPTER 7. TECHNIQUES OF INTEGRATION
518 44.
sec2 x ln(tan x) dx
45.
cos
46. 47.
x dx
√ cos x √ dx x
√ t = x, x = t2 , dx = 2t dt = 2 t cos t dt u = t, du = dt; dv = cos t dt, v = sin t = 2t sin t − 2 sin t dt = 2t sin t + 2 cos t + C √ √ √ = 2 x sin x + 2 cos x + C
u=
√
1 x, du = √ dx 2 x
=2
√
x+C
cos2 x sin x dx u = cos x, du = − sin x dx 2 2 = −2 u2 du = − u3 + C = − cos3 x + C 3 3
(cos2 x − sin2 x) dx =
cos 2x dx =
1 sin 2x + C 2
x2 + 2x + 5 dx = (x + 1)2 + 4 dx =
2
x + 1 = 2 tan θ, dx = 2 sec2 θ dθ 2
4 tan θ + 4(2 sec θ dθ) = 4
See Section 7.3, Example 5
50.
cos u du = 2 sin u + C = 2 sin
cos x sin 2x dx = 2
48.
49.
√
sec2 x dx; dv = sec2 x dx, v = tan x u = ln(tan x), du = tan x = tan x ln(tan x) − sec2 x dx = tan x ln(tan x) − tan x + C
1 (8 − 2x − x2 )3/2
sec3 θ dθ
x2 + 2x + 5
x+1
θ
2 = 2 sec θ tan θ + 2 ln | sec θ + tan θ| + C
√
x2 + 2x + 5 x + 1
1
+ = (x + 1) x2 + 2x + 5 + 2 ln
+C
2 2 2
1
= (x + 1) x2 + 2x + 5 + 2 ln x2 + 2x + 5 + x + 1 + C1 2 1 dx = dx x + 1 = 3 sin θ, dx = 3 cos θ dθ [9 − (x + 1)2 ]3/2 cos θ 1 3 cos θ 1 dθ = sec2 θ dθ = dθ = 3 9 cos3 θ 9 (9 − 9 sin2 θ)3/2 x+1 x+1 1 = tan θ + C = √ +C θ √ 9 9 8 − 2x − x2 8 − 2x − x2
CHAPTER 7 IN REVIEW 51.
5
3
519
tan x sec x dx =
4
2
tan x sec x sec x tan x dx =
(sec2 x − 1)2 sec2 x sec x tan x dx
u = sec x, du = sec x tan x dx 1 2 1 = (u2 − 1)2 u2 du = (u6 − 2u4 + u2 ) du = u7 − u5 + u3 + C 7 5 3 2 1 1 7 5 3 = sec x − sec x + sec x + C 7 5 3 1 x 1 (1 + cos x)2 dx = (1 + 2 cos x + cos2 x) dx 52. cos4 dx = 2 4 4 3 1 1 + cos 2x 1 1 = 1 + 2 cos x + + 2 cos x + cos 2x dx dx = 4 2 4 2 2 1 1 3 sin 2x + C = x + sin x + 8 2 16
53.
t5 dt = 1 + t2
√
54. 55.
t 1 + t2
dt =
1 4 1 2 1 t − t + ln(1 + t2 ) + C 4 2 2
1 dx = sin−1 x + C 1 − x2 x (5x + 1)(x2 + 1) + x 5x + 1 dx + dx = dx 2 2 2 2 (x + 1) x +1 (x + 1)2 1 x 5x dx + dx + dx = 2 2 2 x +1 x +1 (x + 1)2 5 1 +C = ln(x2 + 1) + tan−1 x − 2 2(x2 + 1)
5x3 + x2 + 6x + 1 dx = (x2 + 1)2
√ 56.
t3 − t +
x2 + 9 dx x2
√
x x = 3 tan θ, dx = 3 sec2 θ dθ √ sec3 θ 9 tan2 θ + 9 sec3 θ 2 (3 sec dθ = dθ = θ dθ) = sec2 θ − 1 9 tan2 θ tan2 θ sec θ sec θ = sec θ + dθ dθ = ln | sec θ + tan θ| + sec2 θ − 1 tan2 θ cos θ u = sin θ, du = cos θ dθ = ln | sec θ + tan θ| + 2 dθ sin θ 1 1 = ln | sec θ + tan θ| + du = ln | sec θ + tan θ| − + C 2 u u
√
√
2 x2 + 9 1
x + 9 x
+ C = ln
+ − +C = ln | sec θ + tan θ| −
sin θ 3 3
x
√x 2 + 9
= ln x2 + 9 + x − + C1 x
2
+9
x
θ
3
CHAPTER 7. TECHNIQUES OF INTEGRATION
520
x sin2 x dx
u = x sin x, du = (x cos x + sin x) dx; dv = sin x dx, v = − cos x = −x sin x cos x + x cos2 x dx + sin x cos x dx 1 1 2 sin 2x dx = − x sin 2x + x(1 − sin x) dx + 2 2 1 1 1 = − x sin 2x + x2 − x sin2 x dx − cos 2x 2 2 4 1 1 1 Solving for the integral, x sin2 x dx = x2 − x sin 2x − cos 2x + C. 4 4 8 1 58. (t + 1)2 e3t dt u = (t + 1)2 , du = 2(t + 1) dt; dv = e3t dt, v = e3t 3 1 2 (t + 1)e3t dt = (t + 1)2 e3t − 3 3
57.
1 u = t + 1, du = dt; dv = e3t dt, v = e3t 3 1 2 1 1 (t + 1)e3t − e3t dt = (t + 1)2 e3t − 3 3 3 3 2 2 1 = (t + 1)2 e3t − (t + 1)e3t + e3t + C 3 9 27 sin x sin x 59. e sin 2x dx = e (2 sin x cos x) dx = 2 esin x sin x cos x dx u = sin x, du = cos x dx; dv = esin x cos x, v = esin x sin x = 2e sin x − 2 esin x cos x dx = 2esin x sin x − 2esin x + C
ex tan2 ex dx
60.
π/6
61. 0
√
u = ex , du = ex dx = tan2 u du = (sec2 u − 1) du = tan u − u + C = tan ex − ex + C
cos x dx 1 + sin x
u = 1 + sin x, du = cos x dx
√ √ 3/2 1 √ du = 2 u = 2 3/2 − 2 = 6 − 2 u 1 1 √ π/2 1 1 1+ 2 √ . dx = √ ln 62. Using Mathematica we find sin x + cos x 2 −1 + 2 0 1 63. sinh−1 t dt u = sinh−1 t, du = √ dt; dv = dt, v = t t2 + 1 t −1 √ dt = t sinh−1 t − t2 + 1 + C = t sinh t − t2 + 1 =
3/2
CHAPTER 7 IN REVIEW
x cot x2 dx =
64.
8
65. 3
521
1 ln | sin x2 | + C 2
1 dx x x+1
u2 = x + 1, 2u du = dx
√
3 3 3 2u 2 1 1 du = du = du − du 2 − 1)u (u (u − 1)(u + 1) u − 1 u + 1 2 2 2 2 3 3 3 4 = ln |u − 1| − ln |u + 1| = ln 2 − ln = ln 3 2 2 2 t+1+2 2 1 2 t+3 dt = dt + +C dt = dt = ln |t + 1| − 66. t2 + 2t + 1 (t + 1)2 t+1 (t + 1)2 t+1
3
=
sec4 3u du = cot12 3u
67.
2
68.
x5
tan12 3u(1 + tan2 3u) sec2 3u du
1 1 tan13 3u + tan15 3u + C 39 45
=
tan12 3u sec2 3u sec2 3u du =
4 + x2
x = 2 tan θ, dx = 2 sec2 θ dθ
x2 + 4 dx
0
π/4
= 0
32 tan5 θ
π/4
= 128
0 π/4 0
π/4
tan4 θ sec2 θ tan θ sec θ dθ
0 π/4
= 128
tan5 θ sec3 θ dθ = 128
0
= 128
4 tan2 θ + 4 2 sec2 θ dθ
(sec2 θ − 1)2 sec2 θ tan θ sec θ dθ (sec6 θ − 2 sec4 θ + sec2 θ) tan θ sec θ dθ
π/4 1 2 1 7 5 3 sec θ − sec θ + sec θ = 128 7 5 3 0 √ √ √ 3 7 5 1 2 1 2( 2) ( 2) ( 2) − + − − + = 128 7 5 3 7 5 3 √ 22 √ 8 256 (11 2 − 4) 2− = 128 = 105 105 105
69. 70.
3 + sin x dx = cos2 x
3 sec2 x dx +
tan x sec x dx = 3 tan x + sec x + C
2 sin x cos x dx u = cos x, du = − sin x dx 5 + cos2 x 2u =− du = − ln(5 + u2 ) + C = − ln(5 + cos2 x) + C 5 + u2
sin 2x dx = 5 + cos2 x
θ
2
x
CHAPTER 7. TECHNIQUES OF INTEGRATION
522 71.
x(1 + ln x)2 dx
2(1 + ln x) dx; u = (1 + ln x)2 , du = x 1 = x2 (1 + ln x)2 − x(1 + ln x) dx 2
dv = x dx, v =
1 x2
1 1 u = 1 + ln x, du = dx; dv = x dx, v = x2 x 2 1 1 2 1 x dx = x (1 + ln x)2 − x2 (1 + ln x) − 2 2 2 1 1 1 = x2 (1 + ln x) ln x − x2 (1 + ln x) + x2 + C 2 2 4 1 1 1 x(1 + cos 2x) dx = x2 + x cos 2x dx 72. x cos2 x dx = 2 4 2
x
ex ee dx
73. 74.
1 u = x, du = dx; dv = cos 2x dx, v = sin 2x 2 1 1 1 1 1 1 1 x sin 2x − sin 2x dx = x2 + x sin 2x + cos 2x + C = x2 + 4 2 2 2 4 4 8 x u = ex , du = ex dx = eu du = eu + C = ee + C √
√ √ √ x+1+ x dx = ( x + 1 − x) dx x+1−x 2 2 = (x + 1)3/2 + x3/2 + C 3 3
1 √ √ dx = x+1− x
2t dt 1 + e t2
u = t2 , du = 2t dt 1 e−u du = du = v = e−u , dv = e−u du 1 + eu e−u + 1 2 1 (−dv) = − ln |v + 1| + C = − ln(e−u + 1) + C = − ln(e−t + 1) + C = v+1 2 76. cos x cos 2x dx = cos x(1 − 2 sin x) dx = cos x dx − 2 sin2 x cos x dx
75.
2 sin3 x + C 3 1 1 1 5 77. dx = dx = sin−1 (5x + 2) + C 2 2 5 5 1 − (5x + 2) 1 − (5x + 2) 1 78. (ln 2x) ln x dx u = ln 2x, du = dx; dv = ln x dx, v = x ln x − x x = (ln 2x)(x ln x − x) − (ln x − 1) dx = (ln 2x)(x ln x − x) − ln x dx + x = sin x −
= (ln 2x)(x ln x − x) − x ln x + 2x + C
CHAPTER 7 IN REVIEW cos x ln | sin x| dx
79.
80.
ln
x+1 x−1
523
cos x dx; dv = cos x dx, v = sin x u = ln | sin x|, du = sin x = sin x ln | sin x| − cos x dx = sin x ln | sin x| − sin x + C
dx =
ln(x + 1) dx −
ln(x − 1) dx
u = ln(x + 1), du =
1 dx; x+1
dv = dx, v = x
u = ln(x − 1), du =
1 dx; x−1
dv = dx, v = x
x x dx − x ln(x − 1) + dx = x ln(x + 1) − x+1 x−1 x+1 x−1+1 x+1−1 = x ln dx + dx − x−1 x+1 x−1 x+1 1 1 = x ln − 1− dx + 1+ dx x−1 x+1 x−1 x+1 − x + ln(x + 1) + x + ln(x − 1) + C = x ln x−1 x+1 + ln(x2 − 1) + C = x ln x−1
CÁLCULO INTEGRAL MATEMÁTICAS 2
Chapter 6
UNIDAD 3 APLICACIONES DE LA INTEGRAL
Applications of the Integral
PROBLEMAS 3.1 6.1 Rectilinear Motion Revisited 1. s(t) =
6 dt = 6t + c;
2. s(t) =
5 = s(2) = 6(2) + c;
(2t + 1) dt = t2 + t + c;
c = −7;
s(t) = 6t − 7
0 = s(1) = 12 + 1 + c = 2 + c;
c = −2;
s(t) = t2 + t − 2 3. s(t) =
(t2 − 4t) dt = √
1 3 t − 2t2 + c; 3
6 = s(3) = −9 + c;
c = 15;
s(t) =
1 3 t − 2t2 + 15 3
5 1 9 (4t + 5)3/2 + c; 2 = s(1) = + c; c = − ; 6 2 2 1 5 s(t) = (4t + 5)2/3 − 6 2 π π 5 1 5 5 5 5. s(t) = −10 cos 4t + = s(0) = − dt = sin 4t + + c; + c = − + c; 6 2 6 4 2 2 4 5 π 5 5 + c = ; s(t) = − sin 4t + 2 2 6 2 2 2 2 2 2 6. s(t) = 2 sin 3t dt = − cos 3t + c; 0 = s(π) = + c; c = − ; s(t) = − cos 3t − 3 3 3 3 3 4. s(t) =
4t + 5 dt =
7. v(t) =
−5 dt = −5t + c;
4 = v(1) = −5 + c;
5 (−5t + 9) dt = − t2 + 9t + c; 2 9 5 2 s(t) = − t + 9t − 2 2
s(t) =
c = 9;
2 = s(1) =
335
v(t) = −5t + 9;
13 + c; 2
9 c=− ; 2
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
336 8. v(t) = s(t) =
6t dt = 3t2 + c;
c = −12;
0 = v(2) = 12 + c;
(3t2 − 12) dt = t3 − 12t + c;
v(t) = 3t2 − 12;
−5 = s(2) = −16 + c;
c = 11;
s(t) = t3 − 12t + 11 9. v(t) =
(3t2 − 4t + 5) dt = t3 − 2t2 + 5t + c;
s(t) =
(t3 − 2t2 + 5t − 3) dt =
s(t) =
1 4 2 3 5 2 t − t + t − 3t + 10 4 3 2
−3 = v(0) = c;
1 3 2 3 5 2 t − t + t − 3t + c; 4 3 2
v(t) = t3 − 2t2 + 5t − 3;
10 = s(0) = c;
1 1 (t − 1)2 dt = (t − 1)3 + c; 4 = v(1) = c; c = 4; v(t) = (t − 1)3 + 4; 3 3 1 1 s(t) = (t − 3)3 + 4 dt = (t − 1)4 + 4t + c; 6 = s(1) = 4 + c; c = 2; 3 12 1 (t − 1)4 + 4t + 2 s(t) = 12
10. v(t) =
11. v(t) =
(7t1/3 − 1) dt =
21 4/3 t − t + c; 4
50 = v(8) = 76 + c;
21 4/3 t − t + 26; 4 21 4/3 9 1 t − t − 26 dt = t7/3 − t2 − 26t + c; s(t) = 4 4 2 9 7/3 1 2 s(t) = t − t − 26t − 48 4 2
c = −26;
v(t) =
0 = s(8) = 48 + c;
c = −48;
12. v(t) =
100 cos 5t dt = 20 sin 5t + c;
−20 = v(π/2) = 20 + c;
v(t) = 20 sin 5t − 40; s(t) = (20 sin 5t − 40) dt = −4 cos 5t − 40t + c;
c = −40;
15 = s(π/2) = −20π + c;
c = 15 + 20π;
s(t) = −4 cos 5t − 40t + 15 + 20π 13. v(t) = 2t − 2 = 2(t − 1) 5 |2(t − 1)| dt = 2 dist. = 0
1 = 2 − t2 + t 2
1
0
+2 0
1
−(t − 1) dt + 2 1 2 t −t 2
5
1
=2
1
5
(t − 1) dt
15 1 1 −0 +2 − − = 17 cm 2 2 2
6.1. RECTILINEAR MOTION REVISITED 14. v(t) = −2t + 4 = −2(t − 2) 6 dist. = | − 2(t − 2)| dt = 2 0
1 = 2 − t2 + 2t 2
2
2 0
+2
0
337
−(t − 2) dt + 2
1 2 t − 2t 2
2
6 2
6
(t − 2) dt
= 2(2 − 0) + 2[6 − (−2)] = 20 cm
15. v(t) = 3t2 − 6t − 9 = 3(t + 1)(t − 3) 4 3 4 2 2 dist. = |13t − 6t − 9| dt = 3 −(t − 2t − 3) dt + 3 (t2 − 2t − 3) dt 0
1 = 3 − t3 + t2 + 3t 3
0
3
+3 0
1 3 t − t2 − 3t 3
4 3
3
20 = 3(9 − 0) + 3 − − (−9) = 34 cm 3
3
16. v(t) = 4t − 64t = 4t(t + 4)(t − 4) 4 5 5 |4t3 − 64t| dt = 4 −(t3 − 16t) dt + 4 (t3 − 16t) dt dist. = 1
1 = 4 − t4 + 8t2 4
1
4
+4 1
17. v(t) = 6π cos πt 3 |6π cos πt| dt = 6 dist. = 1
3/2
= 6(− sin πt)]1
1 4 t − 8t2 4
3/2 1
5 4
4
31 175 − (−64) = 306 cm = 4 64 − +4 − 4 4
−π cos πt dt + 6 5/2
5/2
3
π cos πt dt + 6 3/2
5/2
−π cos πt dt
3
+ 6(sin πt)]3/2 + 6(− sin πt)]5/2
= 6[−(−1) − 0] + 6[1 − (−1)] + 6[0 − (−1)] = 24 cm 18. v(t) = 2(t − 3) 7 |2(t − 3)| dt = 2 dist. = 2
3 2
−(t − 3) dt + 2
7 3
(t − 3) dt
3 7 7 1 2 9 9 1 t − 3t −4 +2 − − +2 =2 = 2 − t2 + 3t = 17 cm 2 2 2 2 2 2 3 19. We first convert mi/h to mi/s: 60 mi/h = 60/3600 mi/s. Then the distance traveled is
2 0
60 60 dt = t 3600 3600
2 = 0
1 60 mi = mi × 5280 ft/mi = 176 ft. 1800 30
20. a(t) = −32; v(0) = 0; s(0) = 144; v(t) = −32 dt = −32t + c; 0 = v(0) = c; v(t) = −32t; s(t) = −32t dt = −16t2 + c; 144 = s(0) = c; s(t) = −16t2 + 144 To find when the ball hits the ground, we solve s(t) = −16t2 + 144 = 0. This gives t = ±3. The ball hits the ground in 3 seconds. Its speed at this time is |v(t)| = | − 96| = 96 ft/s.
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
338
21. a(t) = −32; v(0) = 0; s(4) = 0; v(t) = −32 dt = −32t + c; 0 = v(0) = c; v(t) = −32t; s(t) = −32t dt = −16t2 + c; 0 = s(4) = −256 + c; c = 256; s(t) = −16t2 + 256 The height of the building is s(0) = 256 ft. 22. Let the depth of the well be h.
a(t) = −32; v(0) = 0; s(0) = h; v(t) = −32 dt = −32t + c; 0 = v(0) = c; v(t) = −32t; s(t) = −32t dt = −16t2 + c; h = s(0) = c; s(t) = −16t2 + h If tr is the time for the rock to hit the water, then 0 = s(tr ) = −16t2r + h, and h = 16t2r . Since the speed of sound is 1080 ft/s and the sound is heard after 2 seconds, h = 1080(2 − tr ). Then 16t2r = 1080(2 − tr ) or 2t2r + 135tr − 270 = 0. Using the quadratic formula to find the positive root, we obtain √ √ −135 20, 385 −135 + 18, 225 + 2, 160 = ≈ 1.9440 s. tr = 4 4 Then the depth of the well is h = 1080(2 − tr ) ≈ 60.4669 ft. 23. a(t) = −9.8; v(0) = 24.5; s(0) = 0; v(t) = −9.8 dt = −9.8t + c; 24.5 = v(0) = c; v(t) = −9.8t + 24.5; s(t) = (−9.8t + 24.5) dt = −4.9t2 + 24.5t + c; 0 = s(0) = c; s(t) = −4.9t2 + 24.5t Solving v(t) = −9.8t + 24.5 = 0, we see that the maximum height is attained when t = 2.5 seconds. The maximum height is s(2.5) = 30.625 m. 24. a(t) = −3.7; v(0) = 24.5; s(0) = 0; v(t) = −3.7 dt = −3.7t + c; 24.5 = v(0) = c; v(t) = −3.7t + 24.5; s(t) = (−3.7t + 24.5) dt = −1.85t2 + 24.5t + c; 0 = s(0) = c; s(t) = −1.85t2 + 24.5t Solving v(t) = −3.7t + 24.5 = 0 we see that the maximum height is attained when t ≈ 6.6216 seconds. The maximum height is s(6.6216) ≈ 81.1149 m. 25. a(t) = −32; v(0) = 32; s(0) = 384; v(t) = −32 dt = −32t + c; 32 = v(0) = c; v(t) = −32t + 32; s(t) = (−32t + 32) dt = −16t2 + 32t + c; 384 = s(0) = c; s(t) = −16t2 + 32t + 384 Solving v(t) = −32t + 32 = 0 we see that the maximum height is attained when t = 1 second. The maximum height is s(1) = 400 ft. Setting s(t) = −16t2 + 32t + 384 = 0, we have t2 − 2t − 24 = (t − 6)(t + 4) = 0. Thus, the ball hits the ground at 6 seconds.
6.1. RECTILINEAR MOTION REVISITED
339
26. Setting s(t) = −16t2 + 32t + 384 = 256, we have t2 − 2t − 8 = (t − 4)(t + 2) = 0. Thus, the ball passes the observer at 4 seconds. At this time v(4) = −96 ft/s. 27. a(t) = −32; v(0) = −16; s(0) = 102; v(t) = −32 dt = −32t + c; −16 = v(0) = c; v(t) = −32t − 16; s(t) = (−32t − 16) dt = −16t2 − 16t + c; 102 = s(0) = c; s(t) = −16t2 − 16t + 102 Solving s(t) = −16t2 − 16t + 102 = 6, we see that the marshmallow hits the person at t = 2 seconds. The impact velocity is v(2) = −80 ft/s. 28. a(t) = −32; v(0) = 96; s(0) = 22; v(t) = −32 dt = −32t + c; 96 = v(0) = c; v(t) = −32t + 96; s(t) = (−32t + 96) dt = −16t2 + 96t + c; 22 = s(0) = c; s(t) = −16t2 + 96t + 22 Solving s(t) = −16t2 + 96t + 22 = 102, we see that the stone hits the culprit at t = 1 second (or t = 5 seconds if it misses on the way up and hits on its way back down). The impact velocity is v(1) = 64 ft/s. 29. We measure upward from the top of the volcano, so that s(0) = 0. From a(t) = g = −1.8 we obtain v(t) = −1.8t + v0 and s(t) = −0.9t2 + v0 t. If the rock attains its maximum height at time t1 , then v(t1 ) = 0 = −1.8t + v0 and t1 = v0 /1.8. Solving 200, 000 = −0.9t21 + v0 t1 = −0.9 gives v0
v 2 0
1.8
+ v0
v 0
1.8
= 0.9
v 2 0
1.8
=
v02 3.6
3.6(200, 000) ≈ 848.5 m/s.
30. (a) Taking a(t) = −32, v(0) = −2, and s(0) = 25, we have v(t) = −32t−2 25 = and s(t) = −16t2 − 2t + 25. Using similar triangles, we obtain s x 750 , 25(x − 30) = sx and x = . Then x − 30 25 − s
s 30
x – 30 x
dx 750 750 ds 750 = = v(t) = (−32t − 2) dt (25 − s)2 dt (25 − s)2 (25 − s)2 1500(16t + 1) 1500(16t + 1) 375(16t + 1) =− =− =− 2 . (25 − s)2 (16t2 + 2t)2 t (8t + 1)2 375(8 + 1) dx =− 1 (b) When t = 1/2, = −540 ft/s. 2 dt 4 (4 + 1) dv dv dv = v, and integrating with respect to s gives a ds = v 31. From the hint, a = ds. dt ds ds 1 Then as = v 2 + c, and solving for v we have v 2 = 2as − 2c. Since v = v0 when s = 0, 2 v02 = −2c and v 2 = 2as + v02 .
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
340
32. Let a be the acceleration due to gravity, v(0) = v0 , and s(0) = 0. v(t) = a dt = at + c; v0 = v(0) = c; v(t) = at + v0 ; 1 1 s(t) = (at + v0 ) dt = at2 + v0 t + c; 0 = s(0) = c; s(t) = at2 + v0 t 2 2 2v0 1 . Then v(−2v0 /a) = −v0 , and Solving s(t) = at2 + v0 t = 0, we obtain t = 0 and t = − 2 a the speed at impact with the ground is the initial velocity v0 . 33. Let a be the acceleration of gravity on the earth, v(0) = v0 , and s(0) = 0. v(t) = a dt = at + c; v0 = v(0) = c; v(t) = at + v0 ; 1 1 s(t) = (at + v0 ) dt = at2 + v0 t + c 0 = s(0) = c; s(t) = at2 + v0 t 2 2 To find the maximum height reached on earth, we solve v(t) = at + v0 = 0. The maximum height is reached when t = −v0 /a and is s(−v0 /a) = v02 /2a − v02 /a = −v02 /2a. On the planet, the acceleration of gravity is a/2. Proceeding as on the earth, we obtain v(t) = 1 1 at + v0 , and s(t) = at2 + v0 t. To find the maximum height reached on the planet, we 2 4 1 solve v(t) = at + v0 = 0. The maximum height is reached when t = −2v0 /a and is 2 s(−2v0 /a) = v02 /a − 2v02 /a = −v02 /a. Thus, the maximum height reached on the planet is twice that reached on earth. 34. Let a be the acceleration due to gravity on earth. Then, with initial velocity 2v0 , we have 1 ve (t) = at + 2v0 and se (t) = at2 + 2v0 t. On the planet, with acceleration due to gravity 2 1 1 a/2 and initial velocity v0 , we have vp (t) = at + v0 and sp (t) = at2 + v0 t. To find the 2 4 2v0 . The maximum height reached on earth, we solve ve (t) = at + 2v0 = 0 and obtain t = − a 2 2 2 4v 2v 2v maximum height is se (−2v0 /a) = 0 − 0 = − 0 . To find the maximum height reached a a a 1 2v0 . The maximum height on the planet, we solve vp (t) = at + v0 = 0 and obtain t = − 2 a 2 2 2 2v v v is sp (−2v0 /a) = 0 − 0 = − 0 . Thus, the maximum height reached on earth is twice a a a that reached on the planet. We want to find the initial velocity ϑ0 on the earth so that the v2 maximum height reached on earth is − 0 , the maximum height reached on the planet. With a 1 initial velocity ϑ0 , we have ve (t) = at + ϑ0 and se (t) = at2 + ϑ0 t. Solving ve (t) = at + ϑ0 = 0 2 ϑ2 v2 ϑ0 ϑ2 ϑ2 we obtain t = − . Then, we want s(−ϑ0 /a) = 0 − 0 = − 0 to be equal to − 0 . Solving a 2a a√ 2a a for ϑ0 we see that the initial velocity on earth must be 2v0 .
6.2
Area Revisited
PROBLEMAS 3.2 6.2. AREA REVISITED
1
1. A = −1
341
2
−(x − 1) dx =
1 − x3 + x 3
1 −1
2 2 4 = − − = 3 3 3
2 1 -2
-1 1
1
2
2
3
-1 1 -2
1 2 2 1 3 1 x −x −(x2 − 1) dx + (x2 − 1) dx = − x3 + x + 3 3 0 1 0 1 2 2 2 −0 + − − = =2 3 3 3
1
3
2. A =
3. A =
0
1 −x3 dx = − x4 4 −3
0 −3
81 81 =0− − = 4 4
2 1 -1
1 -1
10 -3 3
-2
-1
1
2
-10 -20 -30
1
2
(1 − x3 ) dx + −(1 − x3 ) dx = 0 1 3 3 7 −0 + 2− − = = 4 4 2
4. A =
3
5. A = 0
−(x2 − 3x) dx =
3 1 − x3 − x2 3 2
1 x − x4 4
1 + 0
1 −x + x4 4
2 -1
1
2
1 -5
3 = 0
3
9 9 −0= 2 2
3
-3
0
6. A =
(x + 1)2 dx =
−1
1 (x + 1)3 3
0 = −1
1 1 −0= 3 3
-2
-1
1
-3
0
7. A = −1
3
(x − 6x) dx +
1 0
6
3
−(x − 6x) dx
0 1 1 4 1 x − 3x2 = + − x4 − 3x2 4 4 −1 0 11 11 11 −0 = = 0− − + 4 4 2 1 2 8. A = (x3 − 3x2 + 2) dx + −(x3 − 3x2 + 2) dx 0
-2
-1
1
2
-3 -6
-1
1
1
1 2 1 4 1 x − x3 + 2x + − x4 + x3 − 2x 4 4 1 0 5 5 5 −0 + 0− − = = 4 4 2
=
3
-5
2
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
342
1
−(x3 − 6x2 + 11x − 6) dx +
9. A = 0
3
+ 2
2 1
(x3 − 6x2 + 11x − 6) dx
2
4
-2 -4
−(x3 − 6x2 + 11x − 6) dx
-6
1 2 1 4 11 11 1 x − 2x3 + x2 − 6x + − x4 + 2x3 − x2 + 6x 4 2 4 2 0 1 3 1 4 11 + − x + 2x3 − x2 + 6x 4 2 2 9 9 9 11 − 0 + (−2) − − −2 = = + 4 4 4 4
=
0
3
10. A = −1
(x − x) dx +
1
1
−(x3 − x) dx
0
1 1 4 1 2 1 1 x − x = + − x4 + x2 4 2 4 2 −1 0 1 1 1 −0 = = 0− − + 4 4 2 1 3 1 3 1 1 11. A = −(1 − x−2 ) dx + (1 − x−2 ) dx = −x − + x+ x 1/2 x 1 1/2 1 10 11 5 −2 = = −2 − − + 2 3 6
2
12. A = 1
0
(1 − x
−2
) dx =
1 x+ x
2 = 1
-1
1
-1
2
-2
1 5 −2= 2 2
2
-2
1
13. A = 0
=
4
14. A = 0
=
− 1) dx +
2 − x3/2 + x 3
−(x
1/2
(2 − x
1/2
1
1
+
0
9
) dx +
2 2x − x3/2 3
4
4 + 0
4
2
(x1/2 − 1) dx
2 3/2 x −x 3
4
=
1
4 1 1 −0 + − − =2 3 3 3
4
-2
2
−(2 − x1/2 ) dx
2 −2x + x3/2 3
2
9 = 4
8 8 16 −0 + 0− − = 3 3 3
4
-2
8
6.2. AREA REVISITED
343
0 3 3 4/3 3 4/3 15. A = −x dx + x dx = − x + x 4 4 −2 0 −2 0 3 4/3 3 4/3 3 4/3 = 0 + (2 ) + (3 ) − 0 = (2 + 34/3 ) 4 4 4
0
8
16. A = −1
1/3
3
1/3
(2 − x1/3 ) dx =
3 2x − x4/3 4
2
3 -2
11 27 =4− − = 4 4
8 −1
3
4
0
− sin x dx +
17. A = −π
π 0
0
π
sin x dx = cos x]−π − cos x]0
8
1
-
= [1 − (−1)] − (−1 − 1) = 4
-1
3π
18. A = 0
3
3π
(1 + cos x) dx = (x + sin x)]0 = 3π − 0 = 3π
2 1
2
3
-1
π/2
19. A = −3π/2
−(−1 + sin x) dx = (x +
π/2 cos x)]−3π/2
3π π = − − = 2π 2 2
2
-2
π/3
20. A = 0
sec2 x dx = tan x]0
π/3
=
√
3−0=
√
4
3
2
-
0
21. A = −2
−x dx +
1 0
1 x2 dx = − x2 2
0 + −2
1 3 x 3
1 0
= −(0 − 2) +
1 −0 3
=
7 3
2
-2 2
2
-2
−2
22. A = −3
−(x + 2) dx + −2
√
0
(x + 2) dx + −2
0
0
2
(2 − x2 ) dx +
2 √
√
2
2
2
−(2 − x2 ) dx
1 2 1 2 1 x + 2x x + 2x + + 2x − x3 =− 2 2 3 −3 −2 0 2 1 3 − 2x − x √ 3 2 4 4√ 4√ 3 7 8√ − = − −2 + 2−0 − 2 = + 2 + (0 + 2) + 2 3 3 3 6 3
-2
2
-2
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
344
3
23. A = 0
[x − (−2x)] dx =
3 0
3 3x dx = x2 2
3 = 0
27 2
3
-4
4 -3 -6
2
24. A = 0
(4x − x) dx =
2
3 2 x 2
3x dx = 0
2
9
=6
6
0
3 -4
2
2
25. A = −2
(4 − x ) dx =
1 4x − x2 3
4
16 16 32 − − = = 3 3 3
2 −2
6
3 -4
1
26. A = 0
(x − x2 ) dx =
1 2 1 3 x − x 2 3
1 = 0
4
1 6
2
1
-1
2
27. A = −1
(8 − x3 ) dx =
1 8x − x4 4
1
33 81 = 12 − − = 4 4
2 −1
5
-2
1
28. A =
(x 0
1/3
3
− x ) dx =
3 4/3 1 4 x − x 4 4
1 = 0
2
1 2
1
-1
1 -1
1
29. A = −1
[4(1 − x2 ) − (1 − x2 )] dx =
1 −1
(3 − 3x2 ) dx = (3x − x3 )
1
4
−1 2
= 2 − (−2) = 4 -2
1
30. A = −1
[2(1 − x2 ) − (x2 − 1)] dx =
1 −1
(3 − 3x2 ) dx = (3x − x3 )
= 2 − (−2) = 4
3
31. A = 1
(x − x−2 ) dx =
2
1
3
−1 -2
1 2 1 x + 2 x
3 = 1
10 29 3 − = 6 2 3
2
3
3
6.2. AREA REVISITED
345
9 9 2 3/2 1 1/2 −1/2 1/2 x − 2y 32. A = y− √ (y −y ) dy = dy = y 3 1 1 1 4 40 = 12 − − = 3 3 1 1 2 2 33. A = [(−x + 6) − (x + 4x)] dx = (6 − 4x − 2x2 ) dx
9
√
−3
9 6 3
2
4
−3
1
-3
64 10 − (−18) = 3 3 −3 3/2 3/2 34. A = [(−x2 + 3x) − x2 ] dx = (3x − 2x2 ) dx =
2 6x − 2x2 − x3 3
0
=
3 2 2 3 x − x 2 3 8
35. A = −8
4
3
=
-4
2 1
0
3/2 = 0
(4 − x2/3 ) dx =
9 8
-1
1
2
-1
3 4x − x5/3 5
8 = −8
64 64 128 − − = 5 5 5
4
-8
1
36. A = −1
[(1 − x
2/3
6 2x − x5/3 5
) − (x
1
2/3
− 1)] dx =
4 4 = − − 5 5
1 −1
2
(2 − 2x2/3 ) dx -2
8 = = 5 −1 5 6 37. A = [(2x + 2) − (x2 − 2x − 3)] dx + [(x2 − 2x − 3) − (2x + 2)] dx
−1 5
= −1
(5 + 4x − x2 ) dx +
6 5
(x2 − 4x − 5) dx
6 5 1 3 1 x − 2x2 − 5x 5x + 2x2 + x3 + 3 3 −1 5 8 100 100 118 − − = + (−30) − − = 3 3 3 3 5/2 5/2 5 3 x − x2 dx 38. A = (−x2 + 4x) − x dx = 2 2 0 0 5/2 5 2 1 3 125 x − x = = 4 3 48 0 2 0 1 (x + 6 − x3 ) dx x + 6 + x dx + 39. A = 2 −4 0 0 2 3 2 1 2 1 4 x + 6x x + 6x − x = + = (0 + 12) + (10 − 0) = 22 4 2 4 −4 0
2
-2
20 15
5
8
10 5 -2
2
4
6
=
4 2
-2
2
4
-2
8 6 4 2 -4
-2
2
4
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
346
1
1 y dy = y 3 3 2
40. A = 0
1 = 0
1 3
2 1
-1
1
2
-1
2
41. A = −1
=
1 1 2y + y 2 − y 3 2 3
√
42. A =
[(2 − y 2 ) − (−y)] dy =
3
√ − 3
2
2 −1
0 −2
=
[(6 − y ) − y ] dy =
4
44. A = 0
=
1 −1
=
= 0
0
46. A = −1
8 3
=
0 −2
√3
1
0
0
0
3
(−2y 2 − 4y) dy
2
4 0
(8y − 2y 2 ) dy
4
4
-8
-4
-4
1 −1
(4 + 2x − x3 ) dx
4
13 19 − − = =8 4 4
2
-2
2
2
−(y 3 − y) dy
-2
2
-2
2
π/4 π/4 cos x)]0
6
-2
1 1 1 4 1 2 1 1 = + − y + y =0− − + −0= 4 2 4 4 2 −1 0 π/4 π/2 47. A = (cos x − sin x) dx + (sin x − cos x) dx 1 4 1 2 y − y 4 2
2
3
√ − 3
8 3
[(x + 4) − (x3 − x)] dx =
(y 3 − y) dy +
2 6y − y 3 3
64 3
−1
1
-2
=0− −
1
-1 -1
(6 − 2y ) dy =
2
1 4x − x2 − x4 4
√ − 3
-2
2
[(−y + 2y + 1) − (y − 6y + 1)] dy =
45. A =
3
0
4
1
-3
2
2 4y 2 − y 3 3
√
[(−y − 2y + 2) − (y + 2y + 2)] dy =
−2
2
(2 + y − y 2 ) dy
2
2 − y 3 − 2y 2 3
−1
2
2
43. A =
2
7 10 9 − − = = 3 6 2
√ √ √ = 4 3 − (−4 3) = 8 3
π/2
+ (− cos x − sin x)]π/4 = (sin x + √ √ √ = 2 − 1 + (−1) − (− 2) = 2 2 − 2
-2
6.2. AREA REVISITED
π/2
48. A = 0 2
347
[2 sin x − (−x)] dx =
π/2
(2 sin x + x) dx = 0
1 −2 cos x + x2 2
2
π/2
2
0
16 + π π − (−2) = = 8 8 5π/6 5π/6 49. A = (4 sin x − 2) dx = (−4 cos x − 2x)]π/6
-2
4
π/6
√ √ 5π √ π 12 3 − 4π − −2 3 − =2 3− = 3 3 3 π/2 π/2 50. A = [2 cos x − (− cos x)] dx = 3 cos x dx −π/2
=
−π/2
π/2 3 sin x]−π/2
51. Region 1: y =
√
2
2
-
= 3 − (−3) = 6
-2
4
x, y = −x, x = 0, x = 4 -2
2
4
6
2
4
6
2
4
6
2
4
6
-4
√ Region 2: y = − x, y = x, x = 0, x = 4
4
-2
-4
52. Region 1: y =
1 2 x , y = x − 3, x = −1, x = 2 2
4
-2
-4 4
1 Region 2: y = 3 − x, y = − x2 , x = −1, x = 2 2
4
-2
-4
2
53. 0
1/2 2 3 3 3 − 4x dx + 4x − dx x + 1 − 4x dx = x+1 x+1 0 1/2
1/2
2 = 3 ln |x + 1| − 2x2 0 + 2x2 − 3 ln |x + 1| 1/2 1 3 3 1 = 3 ln − + 8 − 3 ln 3 − + 3 ln 2 2 2 2 3 = 7 + 3 ln ≈ 6.1370 4
6 4 2 -2
2
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
348
1
54. −1
3
55. 0
x e − 2e−x dx =
ln
√
2
2e
−x
−1
−e
x
ln √2
1
dx + ln
√
2
ex − 2e−x dx
4
1 = −2e − e −1 + e + 2e−x ln √2 √ √ √ √ = − 2 − 2 + 2e + e−1 + e + 2e−1 − 2 − 2 √ = 3e + 3e−1 − 4 2 −x
x
-2
x
2
1 9π 9 − x2 dx = π(3)2 = 4 4
4
-3
5
56. −5
6
25 − x2 dx =
(1 + −2
1 25π π(5)2 = 2 2 -5
2
57.
3
5
4
1 4 − x2 ) dx = 1 dx + 4 − x2 dx = 4 + π(2)2 = 4 + 2π 2 −2 −2 2
2
2
-2
1
58. −1
(2x + 3 −
1 − x2 ) dx =
1 −1
(2x + 3) dx −
1
1 −1
4
1 − x2 dx
1 = x2 + 3x −1 − π(1)2 2 π π = 1 + 3 − 1 − 3(−1) − = 6 − 2 2
59. The area of the ellipse is four times the area in the first quadrant portion of the ellipse. Thus, a 4b a 2 4b 1 2 πa = πab. b2 − b2 x2 /a2 dx = a − x2 dx = A=4 a 0 a 4 0
2
-2
2
b
a
6.2. AREA REVISITED 3 1 1 1 1 x+ x+ (−2x + 8) − dx + dx 2 2 2 2 1 2 2 3 5 5 15 5 x− − x+ = dx + dx 2 2 2 2 1 2 2 3 5 2 5 5 2 15 x − x + − x + = 4 2 4 2 1 2 5 5 45 − 10 = =0− − + 4 4 2 −2 −2 √ √ 61. A = (2 − −x − 2) dx + [− −x − 2 − (−2)] dx
2
60. A =
349
(3x − 2) −
−6
0 −2
−2
=2 −6
(2, 4)
(3, 2)
2 (1, 1)
2
−6
+
4
(2 −
[2 − (−2)] dx +
√
−x − 2) dx +
2
0 0
4
4
[2 − (2x − 2)] dx
2
4 dx + −2
0
-6
(4 − 2x) dx
−2 -4
2 2 0 = 2 2x + (−x − 2)3/2 + 4x ]−2 + (4x − x2 ) 0 3 −6 20 52 = 2 −4 − − +8+4−0= 3 3 2 2 2 1 3 1 2 1 1 2 2 y + 1 − (−y − 2) dy = y + y + 3y 62. A = y + y + 3 dy = 2 2 3 4 −2 −2 −2 23 29 52 − − = = 3 3 3 ln 3/2 ln 2 x 63. The area with respect to x is Ax = (e − 1) dx + (2 − ex ) dx. 0 2
ln 3/2
y+1 The area with respect to y is Ay = ln y − ln dy. 2 1 If integration with respect to x is chosen, we get ln 3/2 ln 2 ln 3/2 ln 2 (ex − 1) dx + (2 − ex ) dx = (ex − x)]0 + (2x − ex )]ln 3/2 Ax =
0
ln 3/2
3 3 3 3 = − ln − 1 + 2 ln 2 − 2 − 2 ln + = −3 ln 3 + 5 ln 2 ≈ 0.1699. 2 2 2 2 If integration with respect to y is chosen, we get 2 2 y+1 y+1 + (y + 1) Ay = ln y − ln dy = y ln y − y − (y + 1) ln 2 2 1 1 2 y+1 3 + 1 = 2 ln 2 − 3 ln + 1 − ln 1 + 2 ln 1 − 1 = y ln y − (y + 1) ln 2 2 1 = −3 ln 3 + 5 ln 2 ≈ 0.1699
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
350
(see Problem 1.3.39 5.1.39 for the antiderivative of ln x) 64. Using Mathematica the numbers at which the curves intersect are approximately −0.4077767094044803 The area is then
and
0.7148059123627778
A= −0.4077767094044803
0.7148059123627778.
(ex − 4x2 ) dx ≈ 0.801284.
65. At P (x0 , 1/x0 ) the slope of the line segment is −1/x20 . The equation of the line through Q and R is then y = −x/x20 + 2/x0 . Setting y = 0 we see that the x-intercept is 2x0 . The area is 2x0 2x0 2 2 1 1 2 A= = −2 + 4 = 2, − 2x + dx = − 2 x + x x0 x0 2x0 x0 0 0 which does not depend on x0 . b b 1 2 1 2 1 Ax + Bx Aa + Ba 66. A = (Ax + B) dx = = Ab2 + Bb − 2 2 2 a a A A 2 = (b − a2 ) + B(b − a) = (b + a) + B (b − a) 2 2 f (a) + f (b) Aa + B + Ab + B (b − a) = (b − a) = 2 2
f (b) f (a) a
b
67. By symmetry with respect to the line y = x, a a 1 (cos x − x) dx = 2 sin x − x2 = 2 sin a − a2 A=2 2 0 0 (Using Mathematica it is easily shown that a ≈ 0.739085.) 68. The areas are the same. In Figure 6.2.16(b), 3.2.16(b), the area of the straight swath of paint is k(b − a). 3.2.16(a), then an equation Now, if y = f (x) describes the lower edge of the swath in Figure 6.2.16(a), for the upper edge is y = f (x) + k. The area between the two graphs is then b b {[f (x) + k] − f (x)} dx = k dx = k(b − a). a
a
69. The areas are the same. Let w be the length of the line segments AB and CD, and without loss of generality, let AB reside on y = 0, with CD residing on y = h. Thus, in Figure 6.2.17(a), 3.2.17(a), the area of the rectangle is wh. Since Figure 3.2.17(b), 6.2.17(b) describes a parallelogram, the line defined by AD can be written as x = f (y). Thus, the line defined by BC is x = f (y) + w. The area of the parallelogram is therefore h h {[f (y) + w] − f (y)} dy = w dy = wh. 0
0
70. This project involves a research report, and thus a preset solution is not applicable. It is noted, however, that Cavalieri’s Principle relates directly to the situations presented in Problems 68 and 69.
6.3. VOLUMES OF SOLIDS: SLICING METHOD PROBLEMAS 3.3
6.3
351
Volumes of Solids: Slicing Method
√ √ √ 1. x2 + y 2 = 16; y = 16 − x2 ; A(x) = 3y 2 = 3(16 − x2 ) 4 4√ √ 1 3 2 3(16 − x ) dx = 3 16x − x V = 3 −4 −4 √ √ 128 128 256 3 3 − − ft = 3 = 3 3 3 √ 1 1 2. x2 + y 2 = 16; y = 16 − x2 ; A(x) = πy 2 = π 8 − x2 2 2 4 4 1 1 π 8 − x2 dx = π 8x − x3 V = 2 6 −4 −4 64 64 128π 3 − − ft =π = 3 3 3 4
4 3. x = y 2 ; A(x) = 2y(8y) = 16y 2 = 16x; V = 16x dx = 8x2 0 = 128
(x, y) y x y
4
y√3
y
(x, y) y x y
4
2
y
(x, y)
0
y 4 -2
√ √ 3y 2 3 1 4 2 2 2 = (4 − x ) = 3 4 − 2x + x 4. y = 4 − x ; A(x) = 4 4 4 2 2√ √ 1 2 1 3 4 − 2x2 + x4 dx = 3 4x − x3 + x5 V = 4 3 20 −2 −2 √ √ 64 64 128 3 − − = 3 = 15 15 15 √
2
4
-2
2
√3y/2
2
x
y 5
-2
√
4 − x2 ; A(x) = πy 2 − π(12 ) = π(3 − x2 ) √3 √3 1 3 2 V = √ π(3 − x ) dx = π 3x − x √ 3 − 3 − 3 √ √ √ = π[2 3 − (−2 3)] = 4π 3 ft3
y/2
y
1 2 2π (x − 5)2 5. y = − x + 2; A(x) = πy 2 = 5 2 25 5 5 2π 2π 10π 3 2 2 (x − 5) dx = (x − 5) ft V = = 25 75 3 0 0
6. y =
(x, y)
3
1 -3
-3
√3
x
y
2 3
1
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
352
7. x = −y + 3; A(y) = x2 = (y − 3)2 3 3 1 2 3 =9 V = (y − 3) dy = (y − 3) 3 0 0
3
(x, y)
x
3
8. Let b denote the length of one side of the square base. Thus, B = b2 . h h−y b(h − y) Using similar triangles, we have = and x = . Thus, b 2x 2h 2 2 b (h − y) A(y) = (2x)2 = , and h2 h h 2 b (h − y)2 b2 2 2b2 2 y + dy = − y V = b dy h2 h h2 0 0 h 1 b2 2 b2 3 b2 h 2 = hB. = b2 h − b 2 h + = b y − y + 2y h 3h 3 3 0 9. x =
√
y 1 1 1 1 2 π √ 2 2 V =π [1 − ( y) ] dy = π (1 − y) dy = π y − y = 2 2 0 0 0
h –y h
(x, y)
x
y b
B(1, 1)
C
x O
10. y = x2
1
V =π 0
(x2 )2 dx = π
1
x4 dx =
0
π 5 1 π x = 5 5 0
A
B(1, 1)
C
y O
11. y = x2
1
V =π 0
1 (12 − x4 ) dx = π x − x5 5
B(1, 1)
C
1 = 0
A
4π 5 y O
12. x =
√
y 1 √ ( y)2 dy = π V =π 0
1
y dy = 0
π 2 1 π y = 2 2 0
A
B(1, 1)
C
x O
A
6.3. VOLUMES OF SOLIDS: SLICING METHOD 13. x =
√
y 1 1 √ 2 √ V =π (1 − y) dy = π (1 − 2 y + y) dy 0
1 4 = π y − y 3/2 + y 2 3 2
1 0
π = 6
y 1 1 √ √ V =π [12 − (1 − y)2 ] dy = π (2 y − y) dy 0
=π
x O
√
B(1, 1)
C
0
14. x =
353
4 3/2 1 2 y − y 3 2
A
C
B(1, 1)
x
0
1 = 0
5π 6
O
A
9
2
15. y = 9 − x 3 3 V =π (9 − x2 )2 dx = 2π (81 − 18x2 + x4 ) dx −3
0
1 = 2π 81x − 6x3 + x5 5
3 0
1
=π
17. x =
3
1 y
1 2 y −y 2
5 1
1 x
V =π
5
x
1
15 1 − − =π = 8π 2 2
2
2
1 2 1 2 V =π − 1 dy = π (y −2 − 1) dy y 1/2 1/2 1 5 1 π = π −2 − − = π − −y = y 2 2 1/2 18. y =
-3
√
y−1 5 5 V =π ( y − 1)2 dy = π (y − 1) dy
16. x =
y
1296π = 5
1
1
x 2 6.3.17
3
2 3 1 5π 1 1 dx = π − = π − − (−2) = x x 3 3 1/2 1/2 3
2 1
y 1
2
3
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
354 19. y = (x − 2)2 2 V =π (x − 2)4 dx
4
=
0
π (x − 2)5 5
2 0
=
32π 5
2
y 2
4
√
y−1 1 1 √ √ ( y − 1)2 dy = π (y − 2 y + 1) dy V =π
20. x =
0
=π
1
0
1 2 4 3/2 y − y +y 2 3
1 0
x
π = 6
-1
1 21. y1 = 4 − x2 ; y2 = 1 − x2 4 2 2 2 1 2 15 15 2 2 V = 2π (4 − x ) − 1 − x dx = 2π 15 − x2 + x4 dx 4 2 16 0 0 2 3 5 = 32π = 2π 15x − x3 + x5 2 16 0
y1 y2 -2
2
√
1−y 1 1 1 1 π V =π ( 1 − y)2 dy = π (1 − y) dy = π y − y 2 = 2 2 0 0 0
22. x =
4
2
x 2
23. x1 = y; x2 = y − 1 1 2 V =π y 2 dy + π [y 2 − (y − 1)2 ] dy 0
=π
24. y1 = 1;
1 3 y 3
1
1 0
+ π(y − y)
y2 = 2 − x
1
V =π 0
12 dx + π
2
2 1
2 1
=π
1 +2 3
(2 − x)2 dx = π
1 1 = πx]0 + π 4x − 2x2 + x3 3
2
2
7π = 3
x1 – x2 2
2
2
dx + π 0
x1 1
1
=π+π 1
1
1
8 7 − 3 3
(4 − 4x + x2 ) dx
=
4π 3
1
y1
y2 1
2
6.3. VOLUMES OF SOLIDS: SLICING METHOD
355
25. x = y 2 + 1 2 2 2 2 V =π [5 − (y + 1)] dy = π (4 − y 2 )2 dy
0
0
2
=π 0
26. x = y 2
8 1 (16 − 8y 2 + y 4 ) dy = π 16y − y 3 + y 5 3 5
1
V =π −1
2 2
(1 − y ) dy = π
2 1 = π y − y3 + y5 3 5
1
2 0
2
256π = 15
2
0
3 = π 3x − 3x4/3 + x5/3 5
−1
(1 − 2y + y ) dy
8 8 16π − − =π = 15 15 15
0
1 = 0
0
8 1 = π 4y 2 − y 3 + y 4 − y 5 3 5
2 0
1 = π 9y − y 3 3 1 30. y1 = 9 − x2 ; 2
3 −3
2
2–y
1 1
1
0
2
2–x
2
1
1
−3
-1
64π = 15
y 2 + 16 3 2 y 2 + 16 52 − dy = π V =π
29. x =
1
1–x
3π 5
28. x = −y 2 + 2y 2 2
2 2 2 V =π (8y − 8y 2 + 4y 3 − y 4 ) dy 2 − 2 − (−y + 2y) dy = π
5
4
27. y = x1/3 1 1 V =π [(2 − x1/3 )2 − 12 ] dx = π (3 − 4x1/3 + x2/3 ) dx
5–x
1
1
−1
4
2
3
3 −3
x
(9 − y 2 ) dy
3
= π[18 − (−18)] = 36π
y2 = x2 − 6x + 9 = (x − 3)2 2 4 1 2 4 V =π − (x − 3) dx 9− x 2 0 4 3 4 2 3 =π 108x − 63x + 12x − x dx 4 0 4 3 672π = π 54x2 − 21x3 + 3x4 − x5 = 20 5 0
-3
9
y1
6 3
y2 3
6
9
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
356 31. x1 = y + 6; x2 = y 2 3 V =π [(y + 6)2 − y 4 ] dy = π −2
1 1 = π 36y + 6y 2 + y 3 − y 5 3 5
x1
5
3 −2
x2
(36 + 12y + y 2 − y 4 ) dy 612 664 500π − − =π = 5 15 3
3
−2
6
-5
32. x = (y − 1)1/3 9 9 3π 96π 2/3 5/3 (y − 1) V =π (y − 1) dy = = 5 5 1 1
5
-4
33. y = x3 − x 1 3 2 V =π (x − x) dx = π −1
=π
1 7 2 5 1 3 x − x + x 7 5 3
−1
=π
1 7 1 4 x + x +x 7 2
1
1
8 8 16π − − =π = 105 105 105
= π 3x + 4e
36. y = ex
2
V =π 0
=π
−x
x 2
1 − e−2x 2
2
23 9 16π − − =π = 14 14 7
1 2x e −x 2
=π
0
= π 4e
2
0
0
[(e ) − 1 ] dx = π
1
y
1
2
-1
(x6 + 2x3 + 1) dx
35. y = e−x 1 1 V =π [(2 − e−x )2 − 12 ] dx = π (3 − 4e−x + e−2x ) dx 0
4
y
−1
−1
2
(x6 − 2x4 + x2 ) dx
−1
1
-2 1
−1 1
34. y = x3 + 1 1 V =π (x3 + 1)2 dx = π
x
−1
1 1 − e−2 − 2 2
-2
2
2
1
2–y
2
8
2
(e
2x
0
1 4 5 e − 2 2
− 1) dx 6
4
y 2
1 3
6.3. VOLUMES OF SOLIDS: SLICING METHOD 37. y = | cos x| 2π 2 | cos x| dx = π V =π
357
2π
1 + cos 2x dx 2 0 0 2π π 2π π 1 = (1 + cos 2x) dx = = π2 x + sin 2x 2 0 2 2 0
1
y
2
-1
38. y = sec x π/4 π/4 sec2 x dx = π tan x]−π/4 = π[1 − (−1)] = 2π V =π
2
−π/4
y
1
–/4
39. y = tan x π/4 2 tan x dx = π V =π 0
π/4 0
/4
1
2
(sec x − 1) dx = π(tan x −
π/4 x)]0
y
4π − π 2 π =π 1− −0 = 4 4
/4
40. y1 = cos x; y2 = sin x π/4 2 2 (cos x − sin x) dx = π V =π 0
π/4
cos 2x dx = 0
π/4 π π sin 2x = 2 2 0
1
y1 y2 /4
41. The volume of the right circular cylinder is πr2 h. Placing the center of the red circular cylinder’s base in Figure 3.3.19 6.3.19 on the origin, we see that A = πr2 for every slice from y = 0 to h. Thus, the volume V of the cylinder is h h πr2 dy = πr2 y = πr2 h. V = 0
0
42. Take the cross-sections to be rectangles perpendicular to the base of the cylinder and parallel to the diameter. x2 + y 2 = a2 ; y = a2 − x2 √ √ (a) A(x) = 2yz = (2 a2 − x2 )x = 2x a2 − x2 a V = 2x a2 − x2 dx u = a2 − x2 , du = −2x dx
0
0
= a2
2 −u1/2 du = − u3/2 3
0 a2
2 2 = − (0 − a3 ) = a3 3 3
(x, y) a
z=x x
y x y
2y
z z
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
358
√ √ √ √ (b) A(x) = 2yz = (2 a2 − x2 ) 3x = 2 3x a2 − x2 a √ V = 2 3x a2 − x2 dx u = a2 − x2 , du = −2x dx
0
0
= a2
√
− 3u
1/2
z = √3x
z
2y
x
0 √ √ √ 2 3 3/2 2 3 2 3 3 u (0 − a3 ) = a du = − =− 3 3 3 2
z
a
43. (a) Using Mathematica, we obtain with the disk method 1 4π V =π (5a2 + 9b2 + 21c2 + 105d2 + 18ac + 42bd). [P (x)]2 (1 − x2 ) dx = 315 −1 (b) Setting a = −0.07, b = −0.02, c = 0.2, and d = 0.56 we obtain V ≈ 1.32 cubic units. (c) 1
-1
1
(d) Setting a = −0.06, b = 0.04, c = 0.1, and d = 0.54 we obtain V ≈ 1.26 cubic units. 44. (a) Using x = r2 − y 2 and the disk method, we obtain h−r 2 r2 − y 2 dy V =π
−r
h−r
h−r
(0, h – r) –r
r
1 x h (r2 − y 2 ) dy = π r2 y − y 3 3 –r −r −r 1 1 3 2 2 3 = π r (h − r) − (h − r) − −r + r 3 3 1 = πr2 h − πh3 . 3 4 π (b) The weight of the ball is πr3 ρball and the weight of water displaced is (3rh2 −h3 )ρwater . 3 3 4 π ρball 2 = 0.4 we have π(3) (0.4) = (9h2 − h3 ) or Using Archimedes’ principle and ρwater 3 3 h3 − 9h2 + 43.2 = 0. Solving for h we obtain h ≈ 2.5976 in. =π
45. (a) Each eighth of the bicylinder can be sliced into squares whose sides follow the perimeter of a √ quadrant of the cylinders’ base; that is, x2 + y 2 = r2 , one side of the square is y = r2 − x2 , and its area is y 2 = r2 − x2 . Using symmetry, the volume common to the cylinders is thus r r x3 16r3 . (r2 − x2 ) dx = 8 r2 x − = V =8 3 3 0 0 (b) This item involves a research report, and thus a preset solution is not applicable.
6.4. VOLUMES OF SOLIDS: SHELL METHOD
359
PROBLEMAS 3.4
6.4
Volumes of Solids: Shell Method
1. y =
√
x
1
V = 2π 0
√ 4π 5/2 x x x dx = 5
B(1, 1)
C
1 = 0
4π 5
y O
2. x = y 2
1
V = 2π 0
y(1 − y 2 ) dy = 2π
1 2 1 4 y − y 2 4
A
B(1, 1)
C
1 = 0
π 2
x O
3. x = y 2
1
V = 2π 0
(1 − y)y 2 dy = 2π
1 3 1 4 y − y 3 4
A
B(1, 1)
C
1 = 0
x
π 6 O
4. x = y 2
1
V = 2π 0
y · y 2 dy =
π 4 y 2
1 0
A
B(1, 1)
C
=
x
π 2 O
5. y =
√
x
1
V = 2π 0
√
(1 − x) x dx = 2π
2 3/2 2 5/2 x − x 3 5
1 0
A
B(1, 1)
C
8π = 15
y O
6. y =
√
x
1
V = 2π 0
(1 − x)(1 −
√
1
x) dx = 2π
1 2 2 = 2π x − x3/2 − x2 + x5/2 3 2 5
0
1 0
(1 −
√
A
B(1, 1)
C
x−x+x
3/2
) dx y
7π = 15
O
A
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
360 7. x = y
5
V = 2π 0
y · y dy =
2π 3 y 3
5
5 = 0
250π 3
x
5
8. x = 1 − y 1 1 (y + 2)(1 − y) dy = 2π (2 − y − y 2 ) dy V = 2π 0
1 1 = 2π 2y − y 2 − y 3 2 3
1 0
1
0
x
7π = 3
1
-1
-2
9. x =
√
y
3
3
V = 2π 0
4π 5/2 √ y y y dy = 5
√
3 = 0
36π 3 5
2 1
x
-1
10. y = x2
2
V = 2π 0
1
2
4
π 2 x · x2 dx = x4 = 8π 2 0
y 2
11. y = x2
1
V = 2π 0
1 (3 − x)x dx = 2π x − x4 4 2
3
1 0
4
1
3π = 2
y 1
2
3
6.4. VOLUMES OF SOLIDS: SHELL METHOD 12. x1 =
√
√ x2 = − y
y; V = 2π
9 0
√
√
y[ y − (− y)] dy = 2π
9
361
8π 5/2 y 2y 3/2 dy = 5
0
9 0
x1
1944π = 5
x2 5
-3
3
13. y = x2 + 4 2 2 V = 2π x(x2 + 4 − 2) dx = 2π (x3 + 2x) dx 0
= 2π
1 4 x + x2 4
0
2 = 16π
5
0
y
3
14. y = x2 − 5x + 4 4 4 V = 2π x(−x2 + 5x − 4) dx = 2π (−x3 + 5x2 − 4x) dx 1
5 1 = 2π − x4 + x3 − 2x2 4 3
4
1
= 2π 1
32 7 + 3 12
√ √ 15. x1 = 1 + y; x2 = 1 − y 1 √ √ V = 2π y[1 + y − (1 − y)] dy = 2π 0
8π 5/2 y = 5
1 0
2
135π = 6
= 2π
1 4 8 3 x − x + 8x2 4 3
0
4 = 0
128π 3
4
-2
1
x1
2y 3/2 dy
0
x2
8π = 5
0
y
2
16. y = (x − 2)2 4 4 V = 2π (4 − x)[4 − (x − 2)2 ] dx = 2π (x3 − 8x2 + 16x) dx
2
1
2
2
4
4
2
y
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
362 17. x = y 3
1
V = 2π 0
(y + 1)(1 − y 3 ) dy = 2π
1 1 1 = 2π y + y 2 − y 4 − y 5 2 4 5
1 0
1
= 0
(1 + y − y 3 − y 4 ) dy 2
21π 10
-1
18. y1 = x1/3 + 1; y2 = −x + 1 1 V = 2π (1 − x)[x1/3 + 1 − (−x + 1)] dx
0
1
= 2π 0
= 2π 19. y1 = x;
y2 = x 2
V = 2π 0
2
y1
1
y2
(x1/3 + x − x4/3 − x2 ) dx
3 4/3 1 2 3 7/3 1 3 x + x − x − x 4 2 7 3
1
x
1
2
x(x − x ) dx = 2π
2
1 = 0
1 3 1 4 x − x 3 4
41π 42
1 0
1
π = 6
y2
y1
1
20. y1 = x;
y2 = x 2 1
V = 2π
0
= 2π 0
1
1
2
(2 − x)(x − x ) dx (2x − 3x2 + x3 ) dx
1 = 2π x − x + x4 4 2
3
1 = 0
1
0
3 1 = 2π − x5 + x4 5 4
3 0
2
3
π 2
21. y = −x3 + 3x2 3 3 V = 2π x(−x3 + 3x2 ) dx = 2π (−x4 + 3x3 ) dx
y1
y2
0
243π = 10
4
2
y 2
4
6.4. VOLUMES OF SOLIDS: SHELL METHOD
363
22. y = x3 − x 0 0 1 5 1 3 4π 3 V = 2π −x(x − x) dx = 2π − x + x = 5 3 15 −1 −1
1
y -1
23. y1 = 2 − x2 ; y2 = x2 − 2 0 2 2 V = 2π √ −x[2 − x − (x − 2)] dx = 2π − 2
= 2π
1 4 x − 2x2 2
0 √ − 2
2
0 √
− 2
y1
(2x3 − 4x) dx
= 4π
0
1 = 2π 4x2 + 2x3 − x4 2
2
-2 2
24. y1 = 4x − x2 ; y2 = x2 − 4x 4 4 V = 2π (x + 1)[4x − x2 − (x2 − 4x)] dx = 2π (8x + 6x2 − 2x3 ) dx
y2
-2
3
y1
0
4
y2
= 128π
3
6
-3
0
25. x = y 2 − 5y 5 5 1 4 5 3 625π 2 V = 2π y(−y + 5y) dy = 2π − y + y = 4 3 6 0 0
5
x -5
26. x1 = y + 4; x2 = y 2 + 2 2 2 V = 2π y[y + 4 − (y 2 + 2)] dy = 2π (2y + y 2 − y 3 ) dy 1
1 1 = 2π y + y 3 − y 4 3 4 2
2 = 2π 1
1
8 13 − 3 12
19π = 6
2
x1 x2 2
4
6
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
364
27. y1 = x + 6; y2 = x3 2 2 1 3 1 248π x + 3x2 − x5 V = 2π x(x + 6 − x3 ) dx = 2π = 3 5 15 0 0
6
y1
y2 2
28. x1 = 1 − y 2 ; x2 = y 2 √2/2 V = 2π y(1 − y 2 − y 2 ) dy = 2π 0
= 2π
1 2 1 4 y − y 2 2
√2/2 0
√ 0
2/2
1
(y − 2y 3 ) dy
√2/2
π = 4
1
29. y = sin x2 √π/2 √π/2 2 V = 2π x(1 − sin x ) dx = 2π (x − x sin x2 ) dx 0 0 √π/2 1 2 1 π 1 π 2 − 2π x + cos x2 − = 2π = 2π = 2 2 4 2 2 0
30. y = ex
x1 x2
1
y /2
2
1
V = 2π 0
2
x(ex ) dx = 2π
1 x2 e 2
1 0
= πe − π
2
y 2
31. We use the shell method. r r r h 1 2 h 1 h x dx = 2π hx − x3 V = 2π (r − x) = πr2 h hx − x2 dx = 2π r r 2 3r 3 0 0 0 32. The equation of the line through (r1 , h) and (r2 , 0) is x =
1 (r1 − r2 )y + r2 . We use the disk h
6.4. VOLUMES OF SOLIDS: SHELL METHOD
365
method. 2 1 2 2 2 dy = π (r1 − r2 ) y + r2 (r1 − r2 )y + r2 dy V =π h2 h 0 0 h 1 1 (r1 − r2 )2 y 3 + r2 (r1 − r2 )y 2 + r22 y =π 2 3h h 0 1 2 πh 2 2 2 (r + r1 r2 + r22 ) = πh (r1 − 2r1 r2 + r2 ) + r2 (r1 − r2 ) + r2 = 3 3 1
h
1 (r1 − r2 )y + r2 h
33. We use the disk method.
2
h
r (r2 − y 2 ) dy r2 − y 2 )2 dy = π −r −r r 2 1 2 4 = π r2 y − y 3 = π r3 − − r3 = πr3 3 3 3 3 −r
V =π
r
(
√ 1√ 2 34. The equation of the line is y = r − a2 x and the equation of the circle is y = r2 − x2 . a We use the disk method. 2 b √ 2 b 2 r − a2 V =π x dx + π r2 − x2 dx a a a 2 a b b r − a2 1 3 1 3 r 2 − a2 a 2 2 2 2 x x x dx + π (r − x ) dx = π + π r x − =π a2 a2 3 3 0 a 0 a π 2 1 1 π = (r − a2 )a + π br2 − b3 − ar2 − a3 = (3br2 − 2ar2 − b3 ) 3 3 3 3 35. The equation of the ellipse is y = b
1−
x2 . We use the disk method. a2
2 a a x2 x2 1 V =π dx = πb2 b 1− 2 1 − 2 dx = πb2 x − 2 x3 a a 3a −a −a −a 2 2a 2a 4πab − − = πb2 = 3 3 3
a
x2 . Since the solid is symmetric with respect to a2 the x-axis, we will find the volume of the upper hemispheroid and multiply by 2. We use the shell method. 3/2 a a b2 2 a2 b2 2 πa2 b 2 2 x b − 2 x dx = 4π − 2 = V = 2 2π b − 2x a 3b a 3 0
36. The equation of the ellipse is y = b
1−
0
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
366
ω 2 x2 ω 2 r2 . The depth of the liquid below the x-axis is y2 = h − . 2g 2g So the volume is r 2 2 ω x ω 2 r2 +h− V = 2π x dx 2g 2g 0 r 2 ω 3 2hg − ω 2 r2 x + x dx = 2π 2g 2g 0 2 r ω 4 2hg − ω 2 r2 2 = 2π x + x 8g 4g 0
37. y1 =
=
h
y2 r
√
2hg . The r
πω 2 r4 π(2hg/r2 )r4 1 1 = πr2 h − = πr2 h − πr2 h = πr2 h 4g 4g 2 2
PROBLEMAS 3.5 of a Graph 6.5 Length
1. y = 1;
1
s=
√ 1 + 12 dx = 2 2
−1
2. y = 2;
3
s=
√ 1 + 22 dx = 3 5
0
3. y =
3 1/2 x 2 1
s= 0
8 9 1 + x dx = 4 27
9 1+ x 4
4. y = 2x−1/3 8 −2/3 s= 1 + 4x dx = 1
8 1
3/2 1 0
8 = 27
x2/3 + 4 dx = x2/3
8
13 4
3/2
x−1/3
1
2 −1/3 x dx 3 8 8 3 1/2 du = u3/2 = 83/2 − 53/2 ≈ 11.4471 = u 2 5 5 u = x2/3 + 4, du =
ω2 r2 2g
4πhgr2 − 2πω 2 r4 πω 2 r4 πω 2 r4 + = πr2 h − . 4g 4g 4g
ω 2 r2 38. The liquid will touch the bottom of the bucket when y2 = h − = 0, or ω = 2g volume of the liquid is then V = πr2 h −
y1
−1 =
133/2 − 8 ≈ 1.4397 27
x2/3 + 4 dx
6.5. LENGTH OF A GRAPH
367
5. y = 2x(x2 + 1)1/2 4 2 2 s= 1 + 4x (x + 1) dx = 1
=
2 3 x +x 3
1
4 = 1
4
(2x2
+
1)2
4
dx =
(2x2 + 1) dx
1
140 5 − = 45 3 3
6. y = 2(x + 1)3/2 − 1; y = 3(x + 1)1/2 0 0 0 √ 2 3/2 (9x + 10) s= 1 + 9(x + 1) dx = 9x + 10 dx = 27 −1 −1 −1 2 (103/2 − 1) ≈ 2.2684 = 27
1 1/2 1 −1/2 x−1 x − x = 1/2 2 2 2x 4 4 4 4x + x2 − 2x + 1 (x + 1)2 (x − 1)2 dx = dx = dx s= 1+ 4x 4x 4x 1 1 1 4 1 4 x+1 1 4 1/2 1 2 3/2 1 28 8 10 −1/2 1/2 x − = dx = (x + x ) dx = + 2x = = 2 1 x1/2 2 1 2 3 2 3 3 3 1
7. y =
1 2 1 x4 − 1 x − 2 = 2 2x 2x2 4 4 4 4x + x8 − 2x4 + 1 (x4 − 1)2 s= 1+ dx = dx 4 4x 4x4 2 2 4 4 4 (x + 1)2 1 4 2 1 1 3 1 1 253 13 227 −2 x − = dx = (x + x ) dx = − = = 4 4x 2 2 2 3 x 2 2 12 6 24 2
8. y =
1 4x6 − 1 9. y = x3 − 3 = 4x 4x3 3 3 3 16x6 + 16x12 − 8x6 + 1 (4x6 + 1)2 (4x6 − 1)2 s= 1+ dx = dx = dx 6 6 16x 16x 16x6 2 2 2 3 3 6 4x + 1 1 3 1 1 3 −3 4 dx = (4x + x ) dx = x − 2 = 4x3 4 2 4 2x 2 2 1 1457 127 4685 − ≈ 16.2674 = = 4 18 8 288
CHAPTER 6. APPLICATIONS OF THE INTEGRAL
368
1 4x8 − 1 10. y = x4 − 4 = 4x 4x4 2 2 2 8 16x8 + 16x16 − 8x8 + 1 (4x8 + 1)2 (4x − 1)2 s= 1+ dx = dx = dx 16x8 16x8 16x8 1 1 1 2 1 2 1 4 5 1 1 2 4x8 + 1 4 −4 x dx = (4x + x ) dx = + = 4 1 x4 4 1 4 5 3x3 1 56 1 3067 3011 − ≈ 6.2729 = = 4 120 120 480
(4 − x2/3 )1/2 11. y = − ; x1/3 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨2
8
s= 1
1,
4 − x2/3 1+ dx = x2/3
8 1
2 x1/3
dx = 3x2/3
8 1
2 20, 000. This means n > 9998.5. Thus, it will require S9999 to approximate π/4 to four decimal places.
4.7.2 in the text, we must have an+1 = 46. Using Theorem 9.7.2
47. y =
∞
(−1)k+1 xk−1 = −
k=1
∞
(−1)k xk−1 y =
k=1
(x + 1)y = = =
∞ k=2 ∞ k=1 ∞
(k − 1)(−1)k+1 xk−2
k=2
(k − 1)(−1)k+1 xk−1 + k(−1)k+2 xk +
∞
∞
∞
(k − 1)(−1)k+1 xk−2
k=2
k(−1)k+2 xk−1
k=1
k(−1)k xk +
k=1
(x + 1)y + y = = = =
∞
(−1)k xk−1
k=1 ∞ k=1 ∞ k=1 ∞ k=1 ∞ k=1
k(−1)k xk + k(−1)k xk + k(−1)k xk + k(−1)k xk −
∞ k=1 ∞ k=1 ∞ k=2 ∞ k=1
k(−1)k xk−1 −
∞ k=1
(k − 1)(−1)k xk−1 (k − 1)(−1)k xk−1 k(−1)k xk = 0
(−1)k xk−1
CHAPTER 9. SEQUENCES AND SERIES
640
48. Letting y = J0 (x), we have ∞ ∞ 2k(−1)k 2k−1 2k(2k − 1)(−1)k 2k−2 x , y = x . y = 22k (k!)2 22k (k!)2 k=1 k=1 ∞ ∞ ∞ (2k)(2k − 1)(−1)k x2k−1 (2k)(−1)k 2k−1 (−1)k 2k+1 xy + y + xy = + x + x 22k (k!)2 22k (k!)2 22k (k!)2 =
k=1 ∞ k=1 ∞
k=1
2
k 2k−1
(2k) (−1) x 22k (k!)2
∞ (−1)k 2k+1 + x 22k (k!)2 k=0
k=0
∞
(2k)2 (−1)k x2k−1 (−1)k−1 = + x2k−1 2 22(k−1) [(k − 1)!]2 2(k−1) [(k − 1)!]2 (2k) 2 k=1 k=1 =
∞
∞
(−1)k x2k−1 (−1)k x2k−1 − =0 2(k−1) 2 2 [(k − 1)!] 22(k−1) [(k − 1)!]2 k=1 k=1
∞ ∞ xk−1 xk 49. (a) f (x) = = = f (x) (k − 1)! k!
k=1
k=0
(b) ex
50. (a) f (x) =
f (x) =
∞ (−1)k k=0 ∞
(2k)!
x2k , ∞
(−1)k 2k−1 (−1)k+1 2k+1 x x = (2k − 1)! (2k + 1)!
k=1 ∞
=−
k=0
k=0
k
(−1) x2k+1 = −f (x) (2k + 1)!
(b) f (x) = sin x and f (x) = cos x both satisfy f (x) = −f (x). Since sin x is an odd function while cos x is an even function, the function represented is f (x) = sin x.
9.10. TAYLOR SERIES
9.10
641
Taylor Series
PROBLEMAS 4.10
1. f (x) =
1 , 2−x
f (0) =
1 2
f (x) =
1 , (2 − x)2
f (0) =
1 22
f (x) =
2 , (2 − x)3
f (0) =
2 23
f (x) =
2·3 , (2 − x)4
f (0) =
3! 24
.. . f (k) (x) =
k! , (2 − x)k+1
The Maclaurin series is
f k (0) = ∞ k!/2k+1 k=0
2.
k!
k! 2k+1
xk =
∞ k=0
1 k x . 2k+1
f (x) =
1 , 1 + 5x
f (0) = 1
f (x) =
−5 , (1 + 5x)2
f (0) = −5
f (x) =
52 · 2 , (1 + 5x)3
f (0) = 52 · 2
f (x) =
−53 · 2 · 3 , (1 + 5x)4
f (0) = −53 · 3!
.. . (−1)k 5k k! , f k (0) = (−1)k 5k k! (1 + 5x)k+1 ∞ ∞ (−1)k 5k k! k x = The Maclaurin series is (−1)k 5k xk . k! f (k) (x) =
k=0
k=0
CHAPTER 9. SEQUENCES AND SERIES
642 3.
f (x) = ln(1 + x), f (x) =
f (0) = 0
1 , 1+x
f (x) = − f (x) =
f (0) = 1
1 , (1 + x)2
f (0) = −1
2 , (1 + x)3
f (4) (x) = −
f (0) = 2
2·3 , (1 + x)4
f (4) (0) = −3!
.. .
(k − 1)! , f (k) (0) = (−1)k−1 (k − 1)! (1 + x)k ∞ ∞ ∞ (−1)k−1 (k − 1)! k (−1)k−1 (k − 1)! k (−1)k−1 k x = x = x . The Maclaurin series is k! k! k f (k) (x) = (−1)k−1
k=1
4.
f (x) = ln(1 + 2x), f (x) =
f (0) = 0
2 , 1 + 2x
f (x) = − f (x) =
f (0) = 2
22 , (1 + 2x)2
f (0) = −22
23 · 2 , (1 + 2x)3
f (4) (x) = −
k=1
f (0) = 23 · 2
24 · 2 · 3 , (1 + 2x)4
f (4) (0) = −24 · 3!
.. . 2k (k − 1)! , f (k) (0) = (−1)k−1 2k (k − 1)! (1 + 2x)k ∞ ∞ (−1)k−1 2k (k − 1)! k (−1)k−1 2k k x = x . The Maclaurin series is k! k f (k) (x) = (−1)k−1
k=1
k=1
k=1
9.10. TAYLOR SERIES
643
5. f (x) = sin x,
f (0) = 0
f (x) = cos x,
f (0) = 1
f (x) = − sin x,
f (0) = 0
f (x) = − cos x, .. .
f (0) = −1
f (2k+1) (0) = (−1)k f (2k+1) (x) = (−1)k cos x, ∞ (−1)k 2k+1 The Maclaurin series is x . (2k + 1)! k=1
6. f (x) = cos 2x,
f (0) = 1
f (x) = −2 sin 2x,
f (0) = 0
f (x) = −22 cos 2x,
f (0) = −22
f (x) = 23 sin 2x, .. .
f (0) = 0
f (2k) (0) = (−1)k 22k f (2k) (x) = (−1)k 22k cos 2x, ∞ (−1)k 22k 2k x . The Maclaurin series is (2k)! k=0
7.
f (x) = ex ,
f (0) = 1
f (x) = ex , .. .
f (0) = 1
f (2k) (x) = ex ,
f (k) (0) = 1 ∞ 1 k x . The Maclaurin series is k! k=0
8.
f (x) = e−x ,
f (0) = −1
f (x) = −e−x , .. .
f (0) = 1
f (k) (x) = (−1)k e−x ,
f (k) (0) = (−1)k
The Maclaurin series is
∞ (−1)k k=0
k!
xk .
CHAPTER 9. SEQUENCES AND SERIES
644 9. f (x) = sinh x,
f (0) = 0
f (x) = cosh x,
f (0) = 1
f (x) = sinh x, .. .
f (0) = 0
f (2k+1) (x) = cosh x,
f (2k+1) (0) = 1
The Maclaurin series is
∞ k=0
1 x2k+1 . (2k + 1)!
10. f (x) = cosh x,
f (0) = 1
f (x) = sinh x,
f (0) = 0
f (x) = cosh x, .. .
f (0) = 1
f (2k) (x) = cosh x,
f (2k) (0) = 1 ∞ 1 2k x . The Maclaurin series is (2k)! k=0
11.
f (x) = tan x,
f (0) = 0
f (x) = sec2 x = 1 + tan2 x,
f (0) = 1
f (x) = 2 tan x(1 + tan2 x) = 2 tan x + 2 tan3 x,
f (0) = 0
f (x) = 2 + 8 tan2 x + 6 tan4 x,
f (0) = 2
f (4) (x) = 16 tan x + 40 tan3 x + 24 tan5 x,
f (4) (0) = 0
f (5) (x) = 16 + 136 tan2 x + 240 tan4 x + 120 tan6 x,
f (5) (0) = 16
f (6) (x) = 276 tan x + 1246 tan3 x + 1680 tan5 x + 720 tan7 x,
f (6) (0) = 0
f (7) (0) = 276 f (7) (x) = 276 + 4014 tan2 x + 12, 138 tan4 x + 13, 440 tan6 x + 5040 tan8 x, 2 16 272 7 1 2 17 7 x + · · · = x + x3 + x5 + x + ··· The Maclaurin series is x + x3 + x5 + 3! 5! 7! 3 15 315
9.10. TAYLOR SERIES 12.
645
f (x) = sin−1 x,
f (0) = 0
f (x) = (1 − x2 )−1/2 ,
f (0) = 1
f (x) = x(1 − x2 )−3/2 ,
f (0) = 0
f (x) = 3x2 (1 − x2 )−5/2 + (1 − x2 )−3/2 ,
f (0) = 1
f (4) (x) = 15x3 (1 − x2 )−7/2 + 9x(1 − x2 )−5/2 ,
f (4) (0) = 0
f (5) (x) = 105x4 (1 − x2 )−9/2 + 90x2 (1 − x2 )−7/2 + 9(1 − x2 )−5/2 ,
f (5) (0) = 9
f (6) (x) = 945x5 (1 − x2 )−11/2 + 1050x3 (1 − x2 )−9/2 + 225x(1 − x2 )−7/2 ,
f (6) (0) = 0
f (7) (x) = 10, 395x6 (1 − x2 )−13/2 + 14, 175x4 (1 − x2 )−11/2 + 4, 725x2 (1 − x2 )−9/2 + 225(1 − x2 )−7/2 , f 7 (0) = 225 The Maclaurin series is x +
13. f (x) =
1 , 1+x
f (x) = − f (x) =
f (4) =
1 , (1 + x)2
2 , (1 + x)3
f (x) = −
9 225 7 1 3 x + x5 + x + ··· . 3! 5! 7!
1 5
f (4) = − f (4) =
2·3 , (1 + x)4
1 52
3 53
f (4) = −
3! 54
.. . (−1)k k! (−1)k k! , f (2k+1) (4) = k+1 (1 + x) 5k+1 ∞ ∞ (−1)k k!/5k+1 (−1)k (x − 4)k = The Taylor series is (x − 4)k . k! 5k+1 f (2k+1) (x) =
k=0
k=0
CHAPTER 9. SEQUENCES AND SERIES
646 14.
f (x) = x1/2 ,
f (1) = 1
1 −3/2 x , 2
f (x) =
f (x) = −
f (1) =
3 −5/2 x , 22
1 2
f (1) = −
3 22
3·5 3·5 f (x) = 3 x−7/2 , f (1) = 3 2 2 .. . (−1)k+1 1 · 3 · 5 · · · (2k + 1) −(2k+2)/2 (−1)k+1 1 · 3 · 5 · · · (2k + 1) x f (k) (1) = f k (x) = k 2 2k The Taylor series is ∞ ∞ (−1)k+1 1 · 3 · 5 · · · (2k + 1)/2k (−1)k+1 1 · 3 · 5 · · · (2k − 1) (x − 1)k = 1 + (x − 1)k . 1+ k! 2k k! k=0
15. f (x) =
k=0
1 , x
f (x) = − f (x) =
f (1) = 1 1 , x2
2 , x3
f (1) = −1 f (1) = 2
6 f (x) = − 4 , f (1) = −6 x .. . (−1)n n! f (n) (1) = (−1)n n! f n (x) = xn+1 The Taylor series is ∞ ∞ ∞ f ( k)(1) (−1)k k! (x − 1)k = (x − 1)k = (−1)k (x − 1)k k! k! k=0
k=0
k=0
9.10. TAYLOR SERIES 16. f (x) =
647
1 , x
f (x) = − f (x) =
f (−5) = − 1 , x2
1 5
f (−5) = −
2 , x3
1 25
f (−5) = −
2 125
6 6 f (x) = − 4 , f (−5) = − x 125 .. . (−1)n n! n! f n (x) = f (n) (−5) = − n+1 n+1 x 5 The Taylor series is ∞ ∞ f ( k)(−5) −1 (x + 5)k = (x + 5)k k! 5k+1 k=0
k=0
17.
√
f (x) = sin x,
f (π/4) =
f (x) = cos x,
f (π/4) =
f (x) = − sin x,
√ f (π/4) = − 2/2
2/2
√
2/2
√ f (π/4) = − 2/2 √ √ √ √ 2 2 2 2 π π 2 π 3 The Taylor series is + − + ··· x− − x− x− 2 2 4 2 · 2! 4 2 · 3! 4 f (x) = − cos x,
18. f (x) = sin x,
f (π/2) = 1
f (x) = cos x,
f (π/2) = 0
f (x) = − sin x,
f (π/2) = −1
f (x) = − cos x, .. .
f (π/2) = 0
f (2k) (π/2) = (−1)k f (2k) (x) = (−1)k sin x, ∞ (−1)k π 2k x− The Taylor series is . (2k)! 2 k=0
CHAPTER 9. SEQUENCES AND SERIES
648 19. f (x) = cos x,
f (π/3) = 1/2
f (x) = − sin x,
√ f (π/3) = − 3/2
f (x) = − cos x,
f (π/3) = −1/2
√ f (π/3) = 3/2 √ √ 3 3 1 π 1 π 2 π 3 The Taylor series is − + + ···. x− − x− x− 2 2 3 2 · 2! 3 2 · 3! 3 f (x) = sin x,
20.
√
3/2
f (x) = cos x,
f (π/6) =
f (x) = − sin x,
f (π/6) = −1/2
f (x) = − cos x,
√ f (π/6) = − 3/2
f (x) = sin x,
f (π/6) = 1/2 √ 3 1 3 1 π π 2 π 3 − The Taylor series is + + ···. x− − x− x− 2 2 6 2 · 2! 6 2 · 3! 6 21.
√
f (x) = ex ,
f (1) = e
f (x) = ex , .. . f (k) (x) = ex ,
f (1) = e
f (k) (1) = e ∞ e (x − 1)k . The Taylor series is k! k=0
22.
f (x) = e−2x ,
f (1) = e−1
f (x) = −2e−2x ,
f (1) = −2e−1
f (x) = 22 e−2x ,
f (1) = 22 e−1
f (x) = −23 e−2x , .. .
f (1) = −23 e−1
f (k) (1) = (−1)k 2k e−1 f (k) (x) = (−1)k 2k e−2x , k ∞ (−1k 2k e−1 ) 1 The Taylor series is . x− k! 2 k=0
9.10. TAYLOR SERIES
649
23. f (x) = ln x, f (x) =
f (2) = 0
1 , x
f (x) = − f (x) =
f (2) = 1 , x2
1 2
f (2) = −
2 , x3
2 23
f (2) =
2·3 f (4) (x) = − 4 , x .. . f (k) (x) = (−1)k−1
1 22
f (4) (2) = − (k − 1)! , xk
3! 24
f (k) (2) = (−1)k−1
(k − 1)! 2k
The Taylor series is ln 2 +
∞ (−1)k−1 (k − 1)!/2k
k!
k=1
(x − 2)k = ln 2 +
∞ (−1)k−1 k=1
k2k
(x − 2)k .
24. f (x) = ln(x + 1), f (x) =
f (2) = ln 3
1 , x+1
f (x) = − f (x) =
f (2) =
1 , (x + 1)2
f (2) = −
2 , (x + 1)3
f (4) (x) = −
1 3
f (2) =
2·3 , (x + 1)4
1 32
2 33
f (4) (2) = −
3! 34
.. .
(k − 1)! (k − 1)! , f (k) (2) = (−1)k−1 (x + 1)k 3k ∞ ∞ (−1)k−1 (k − 1)!/3k (−1)k−1 (x − 2)k = ln 3 + The Taylor series is ln 3 + (x − 2)k . k! k3k f (k) (x) = (−1)k−1
k=1
25. Substituting x2 for x in Problem 8, we have
k=1
∞ k=0
(−1)k 2 k (x ) = k!
∞ k=0
26. Substituting 3x for x and mutiplying by x2 in Problem 8, we have x2
(−1)k 2k x . k! ∞ (−1)k k=0
k!
(3x)k =
∞ (−1)k 3k k=0
k!
xk+2 .
CHAPTER 9. SEQUENCES AND SERIES
650
27. Multiplying by x in Example 3 in the text, we have x
∞ (−1)k k=0
(2k)!
x2k =
k=0
29. Substituting −x for x in Problem 3, we have
∞ k=1
30. Using Problems 3 and 20 and the fact the ln
k=1
k
xk −
∞ −1 k=1
k
xk =
∞ (−1)k−1 + 1
k
k=1
k=0
(2k)!
x2k+1 .
∞ ∞ (−1)k (−1)k 6k+3 (x3 )6k+3 = x . (2k + 1)! (2k + 1)!
28. Substituting x3 for x in Problem 5, we have
∞ (−1)k−1
∞ (−1)k
k=0
(−1)k−1 (−x)k = k
1+x 1−x
∞ k=1
−1 k x . k
= ln(1 + x) − ln(1 − x), we have ∞
2 2 2 xk = 2x + x3 + x5 + · · · = x2k−1 . 3 5 2k − 1 k=1
d tan x, we have dx d 2 17 7 2 17 1 x + · · · = 1 + x2 + x4 + x6 + · · · . x + x3 + x5 + dx 3 15 315 3 45
31. Using Problem 11 and the fact that sec2 x =
x 32. Using Problem 11 and the fact that ln(cos x) = − 0 tan tdt, we have x 1 2 16 276 8 2 3 16 5 276 7 t + · · · dt = − x2 − x4 − x6 − x − ··· . t+ t + t + − 3! 5! 7! 2 4! 6! 8! 0 x3 = lim x→0 x − sin x x→0
33. lim
x3 3
x5 x + − ··· x− x− 6 120 x3 = lim x3 x5 x→0 6 − 120 + · · · 1 = lim 1 x2 x→0 − 6 120 + · · · 1 = 1 =6
6
x2 x3 x4 1 + x − 1 + x + + + + · · · 2 6 24 1+x−e
= lim 34. lim x2 x4 x→0 1 − cos x x→0 1 − 1 − 2 + 24 − · · · x
2
= lim
x→0
= lim
x→0
x3 x4 6 − 24 − · · · x2 x4 2 − 24 + · · · 2 − x3 − x12 − · · · = 2 1 − x12
− x2 − −1
−1 = −1 1
9.10. TAYLOR SERIES 35. cosh x =
ex + e−x 2 1+x+
=
2 1+
=
x2 2!
651
x2 2!
+
+
x4 4!
x3 3!
+ 1−x+
x2 2!
−
x3 3!
x2 2!
−
x3 3!
2 + ···
2
∞ x2k = (2k)! k=0
36. sinh x =
ex − e−x 2 1+x+
=
2 x+
=
x3 4!
x2 2!
+
+
x5 5!
x3 3!
− 1−x+
2
+ ···
2 ∞ 2k+1 x = (2k + 1)! k=0
1 ex x =e · 37. 1−x 1−x
x3 x4 x2 + + + · · · · 1 + x + x 2 + x3 + x4 + · · · = 1+x+ 2! 3! 4! 5 2 8 3 65 4 = 1 + 2x + x + x + x + · · · 2 3 24
x3 x4 x2 + + 2! 3! 4! 3 5 x x = x + x2 + − − ··· 3 30
38. ex sin x =
1+x+
2
x5 x3 + + ··· ··· x + 3! 5!
3
1 + x + x2 + x3! + · · · ex 39. = 2 4 cos x 1 − x2 + x4! − · · · = 1 + x + x2 +
40. sec x =
1 = cos x 1− 2
=1+
x4 2x3 + + ··· 3 2 1
x2 2 4
+
x4 4!
− 6
x5 5!
+ ···
5x 61x 277x8 x + + + + ··· 2 24 720 8064
CHAPTER 9. SEQUENCES AND SERIES
652
−1
41. 0
1
42. 0
−1
(−x2 )3 (−x2 )2 + + · · · dx 1 + (−x2 ) + 2 3! 0 −1 x6 x4 − + · · · dx = 1 − x2 + 2 6 0 3 5 7 1 x x x = x− + − 3 10 42 0 1 1 1 − + ··· =1− + 3 10 42 1 x5 x7 sin x 1 x3 dx = + − + · · · dx x− x 3! 5! 7! 0 x 1 2 4 6 x x x = + − + · · · dx 1− 3! 5! 7! 0 1 x5 x7 x3 + − = x− 3 · 3! 5 · 5! 7 · 7! 0 1 1 1 + − + ··· =1− 3 · 3! 5 · 5! 7 · 7! 3
e−x dx =
43. Using tan−1 x =
∞ k=0
(−1)k
x2k+1 , 2k + 1
we have ∞
(−1)k π 1 1 1 = tan−1 1 = = 1 − + − + ··· 4 2k + 1 3 5 7 k=0
44. Using ex = 1 + x +
x3 x2 + + · · · , we have 2! 3! 1 1 1 1 − + − + ··· 2! 3! 4! 5! 1 1 1 1 = − + − + ··· 2! 3! 4! 5!
e−1 = 1 − 1 +
45. Using cos x = 1 −
x4 x6 x2 + − + · · · , we have 2! 4! 6! cos π = 1 −
46. Using sin x = x −
π2 π4 π6 + − + ··· 2! 4! 6!
x5 x7 x3 + − + · · · , we have 3! 5! 7! sin π = π −
π3 π5 π7 + − 3! 5! 7!
9.10. TAYLOR SERIES
653
47. Using Problem 17 and the fact that 46◦ ≈ 0.802851456 radians, we have √ √ √ 2 2 2 π π 2 ◦ + ≈ 0.719340424. sin 46 ≈= 0.802851456 − − 0.802851456 − 2 2 4 4 4 Now, for f (x) = sin x, f (x) = − cos x and |R2 (x)| = |R2 (0.802851456)| =
|0.802851456 − π/4|3 < 0.000001. 6
| cos c| |x − π/4|3 |x − π/4|3 < . Thus, 3! 3!
48. Using Problem 20 and the fact that 29◦ ≈ 0.506145483 radians, we have √ √ 3 1 3 π π 2 − ≈ 0.874620147. cos 29◦ ≈ 0.506145483 − − 0.506145483 − 2 2 6 4 6 Now, for f (x) = cos x, f (x) = sin x and |R2 (x)| = |R2 (0.506145483)| <
|0.506145483 − π/6|3 < 0.000001. 6
| sin c| |x − π/6|3 |x − π/6|3 < . Thus, 3! 3!
(0.3)3 (0.3)4 (0.3)2 + + ≈ 1.349837500. Now, 2 3! 4! c 5 e 3|x| 3|0.3|5 |R4 (x)| = |x|5 < since 0 < c < 1 and ec < e < 3. Thus, |R4 (0.3)| < < 0.0001. 5! 5! 5!
49. Using Problem 7, we have e0.3 ≈ 1 + 0.3 +
(0.1)3 ≈ 0.100166667. Now, for f (x) = sin hx, 3! | sinh c| 4 (sinh 1)|x|4 f (4) (x) = sinh x and |R3 (x)| = |x| < since 0 < c < 0.1 and sinh c < 4! 4! −1 2 e−e e −1 9−1 2(0.1)4 sinh 1 = = < = 2. Thus, |R3 (0.1)| < < 0.00001. 2 2e 4 24 | cos x|, n even (n+1) 51. We use |f (x)| = . Since | cos x| ≤ 1 and | sin x| ≤ 1, |Rn (x)| = | sin x|, n odd |x|n+1 |f (n+1) (c)| n+1 |x| and lim |Rn (x)| = 0. Thus, the series represents sin x for all x. ≤ n→∞ (n + 1)! (n + 1)! 50. Using Problem 9, we have sinh(0.1) ≈ 0.1 +
ec |x|n+1 , where c is between 0 and x. If x < 0, then ec < 1 and (n + 1)! ec |x|n+1 ec |Rn (x)| = |x|n+1 < . If x > 0, then ec < ex and |Rn (x)| = |x|n+1 < (n + 1)! (n + 1)! (n + 1)! ex |x|n+1 . In either case, for x fixed, lim |Rn (x)| = 0 and the series represents ex for all x. n→∞ (n + 1)! ⎧ sinh c n+1 ⎪ ⎪ , n odd ⎨ (n + 1)! x 53. We use Rn (x) = for c between 0 and x. Now, | sinh c| < | sinh x| and cosh c n+1 ⎪ ⎪ ⎩ x , n even (n + 1)! | sinh c| n+1 | sinh x||x|n+1 |x| , < cosh c < cosh x for c between 0 and x, so for n odd, |Rn (x)| = (n + 1)! (n + 1)!
52. We use Rn (x) =
CHAPTER 9. SEQUENCES AND SERIES
654
cosh c (cosh x)|x|n+1 |x|n+1 < . Thus, lim |Rn (x)| = 0 and the n→∞ (n + 1)! (n + 1)! series represents sinh x for all x. ⎧ cosh c n+1 ⎪ ⎪ , n odd ⎨ (n + 1)! x 54. We use Rn (x) = for c between 0 and x. Now, | sinh c| < | sinh x| sinh c n+1 ⎪ ⎪ ⎩ x , n even (n + 1)! | sinh c| n+1 |x| < and cosh c < cosh x for c between 0 and x, so for n even, |Rn (x)| = (n + 1)! | sinh x||x|n+1 cosh c (cosh x)|x|n+1 , and for n odd, |Rn (x)| = |x|n+1 < . Thus, lim |Rn (x)| = n→∞ (n + 1)! (n + 1)! (n + 1)! 0 and the series represents cosh x for all x. and for n even, |Rn (x)| =
R+y . This gives y = R sec x − R = 55. (a) From Figure 9.10.3, 4.10.3 L = Rx and sec x = R L R sec −R R (b) We need to find the Maclaurin series for f (x) = sec x. We compute f (x) = sec x,
f (0) = 1
f (x) = sec x tan x,
f (0) = 0
f (x) = 2 sec3 x − sec x,
f (0) = 1
f (x) = 6 sec3 x tan x − sec x tan x,
f (0) = 0
f (4) = 6 sec5 x + 18 sec3 tan2 x − sec3 x − sec x tan x,
f (4) (0) = 5
x2 + · · ·. 2 x2 Approximating sec x by 1 + , we have 2 2 L L L2 (L/R)2 −1 =R y = R sec = −1 =R 1+ R 2 2r2 2R Therefore, sec x = 1 +
(c) Using y ≈
L2 , we have 2R y=
(5280)2 ft ≈ 0.66 ft = 7.92 in 2(4000)(5280)
5 x2 + x4 , we have (d) Approximating sec x by 1 + 2 24 5 (L/R)2 + (L/R)4 − 1 y =R 1+ 2 24 2 4 5L L + = . 2R 24R3
9.10. TAYLOR SERIES
655
56. (a) Since lim tan h d→∞
2πd = 1, for d large, ν ≈ gL/2π. L
(b) f (x) = tanh x, f (0) = 0 f (x) = sech2 x = 1 − tanh2 x, f (0) = 1 f (x) = −2 tanh x(1 − tanh2 x) = 2 tanh3 x − 2 tanh x, f (0) = 0 f (x) = 6 tanh2 x(1−tanh2 x)−2(1−tanh2 x) = −6 tanh4 x+8 tanh2 x−2, f (0) = −2 f (4) (x) = −24 tanh3 x(1 − tanh2 x) + 16 tanh x(1 − tanh2 x), f (4) (0) = 0 = 24 tanh5 x − 40 tanh3 x + 16 tanh x, f (x) = 120 tanh4 x(1−tanh2 x)−120 tanh2 x(1−tanh2 x)+16(1−tanh2 x), f (5) (0) = 2 16 5 1 2 5 16 The Maclaurin series is tanh x = x − x3 + x − · · · = x − x3 + x − ··· . 6 120 3 15 √ Now, for small d/L, tanh(2πd/L) ≈ 2πd/L and ν ≈ (gL/2π)(2πd/L) = gd. (5)
57. sin2 x = (sin x)(sin x) x5 x7 x3 x5 x7 x3 + − + ··· x− + − + ··· = x− 3! 5! 7! 3! 5! 7! 2x6 x4 = x2 − + − ··· 3 45 Also, sin2 x = 1 − cos2 x − 1 − (cos x)(cos x) x4 x2 x4 x2 + − ··· 1− + − ··· =1− 1− 2! 4! 2! 4! 4 6 2x x = 1 − 1 − x2 + − + ··· 3 45 2x6 x4 = x2 − + − ··· 3 45
x3 x5 x7 x2 x4 + − + ··· 1− + − ··· 3! 5! 7! 2! 4! 2x3 2x5 =x− + − ··· 3 15 Also, using the result from Problem have 57, we 1 d 2x6 1 d x4 2 2 sin x cos x = [sin x] = + − ··· x − 2 dx 2 dx 3 45 1 4 4 = 2x − x3 + x5 − · · · 2 3 15 2 2 = x − x3 + x5 − · · · 3 15
58. sin x cos x =
x−
59. Using ex = 1 + x +
x3 x2 + + · · · , we have 2 3!
CHAPTER 9. SEQUENCES AND SERIES
656
(x + 1)2 ex = (x + 1)2 ex+1−1 = e−1 (x + 1)2 ex+1 (x + 1)3 (x + 1)2 + + ··· = e−1 (x + 1)2 1 + (x + 1) + 2 3! 4 (x + 1)5 (x + 1) = e−1 (x + 1)2 + (x + 1)3 + + + ··· 2 3! ∞ −1 k+2 e (x + 1) = k! k=0
60. No, the function f (x) = cot x is undefined at x = 0. 61. cos x is an even function while sin x is an odd function. From (18), (19), and (20), we see that tan−1 x is an odd function, cosh x is an even function, and sinh x is an odd function. (x2 )3 f (x) = x4 sin x2 = x4 (x2 ) − + ··· 3! 10 x = x6 − + ··· 3! f (10) (0) −10! The coefficient of x10 should be . Therefore, f (10) (0) = = −604, 800. 10! 3!
62. We have
63.
PROBLEMAS 4.11 9.11 Binomial Series 1. With r = 1/3, for |x| < 1, √ 3
1 1 1 1 1 1 2 3 3 −1 3 3 −1 3 −2 x + x3 + · · · 1+x=1+ x+ 3 2! 3! 2 1 2·5 3 x2 + 3 x − ··· =1+ x− 2 3 3 · 2! 3 · 3!
. 2. With r = 1/2, for |x| < 1, √
1
1 1 1 1 1 2 2 2 −1 2 2 −1 2 −2 (−x) + (−x)3 + · · · 1 + x = 1 + (−x) + 2 2! 3! 1 1 3 x2 − 2 x3 − · · · . =1− x− 2 2 2 · 2! 2 · 3!
3. With r = 1/2, for |x/9| < 1 or |x| < 9, √ 1 x − + 9 − x = 3 1 − x/9 = 3 1 + 2 9 =3−
1 2
1
x 2 2 −1 − + 2! 9
3·1 3 3·1·3 3 x− 2 x2 − 3 x − ··· . 2 2·9 2 · 2! · 9 2 · 3! · 93
1 2
1 2
− 1 12 − 2 x 3 − + ··· 3! 9
9.11. BINOMIAL SERIES
657
4. With r = −1/2, for |5x| < 1 or |x| < 1/5,
− 12 − 12 − 1 − 12 − 2 − 12 − 12 − 1 1 1 2 √ (5x) + (5x)3 + · · · =1+ − (5x) + 2 2! 3! 1 + 5x 3 · 52 2 3 · 5 · 53 3 5 x − 3 x − ··· . =1− x− 2 2 2 · 2! 2 · 3! 5. With r = −1/2, for |x| < 1,
− 12 − 12 − 1 − 12 − 12 − 1 − 12 − 2 1 1 2 2 2 √ (x ) + (x2 )3 + · · · =1+ − (x ) + 2 2! 3! 1 + x2 3 3·5 6 1 x4 − 3 x − ··· . =1− x− 2 2 2 · 2! 2 · 3! 6. With r = −1/3, for |x| < 1,
1
1 1 1 1 − − − − − 1 − 1 − − 2 1 x 3 3 3 √ (−x2 )2 + 3 (−x2 )3 + · · · = x 1 − (−x2 ) + 3 3 3 2! 3! 1 − x2 1 4 4·7 6 x5 + 3 x + ··· . = x + x3 + 2 3 3 · 2! 3 · 3! 7. With r = 3/2, for |x/4| < 1 or |x| < 4, 3 x 3/2 3/2 = 8(1 + x/4) =8 1+ (4 + x) + 2 4 =8+
x (1 +
3
− 1 x 2 + 2! 4
2
3 2
3 2
− 1 32 − 2 x 3 + ··· 3! 4
8 · 3 · 1 2 8 · 3 · 1 · (−1) 3 8·3 x+ 2 x + x + ··· . 2·4 3 · 2! · 42 23 · 3! · 43
8. With r = −5/2, for |x| < 1, ⎡
3 2
x)5
5 ⎢ = x ⎣1 − x + 2
− 52
− 52 2!
−1
x2 +
−
⎤
5 5 − 2 − 1 − 52 − 2 ⎥ 2 x3 + · · · ⎦ 3!
5·7 3 5·7·9 4 5 x − 3 x + ··· = x − x2 + 2 2 2 · 2! 2 · 3! . 9. With r = −2, for |x/2| < 1 or |x| < 2, x x = (1 + x/2)−2 2 (2 + x) 4 x −2(−2 − 1) x 2 −2(−2 − 1)(−2 − 2) x 3 x + = + + ··· 1−2 4 2 2! 2 3! 2 1 2·3 2·3·4 4 1 x3 − x + ··· = x − x2 + 4 4 4 · 2! · 22 4 · 3! · 23 .
CHAPTER 9. SEQUENCES AND SERIES
658 10. With r = −3, for |x| < 1,
−3(−3 − 1) −3(−3 − 1)(−3 − 2) ((−x)2 )2 + ((−x)2 )3 + · · · x2 (1 − x2 )−3 = x2 1 − 3(−x)2 + 2! 3! 3 · 4 3 · 4 · 5 x6 + x8 + · · · = x2 + 3x4 + 2! 3! . 11. See Problem 1. Since the series is alternating on (0, 1), by Theorem 4.7.2 9.7.2 the approximation x 1 2 to the sum using S2 = 1 + is accurate within a3 = x . 3 9
4.7.2 the approximation 12. See Problem 5. Since the series is alternating on (−1, 1), by Theorem 9.7.2 3 4 x2 3·5 6 5 6 + x is accurate within a4 = 3 x = x . to the sum using S3 = 1 − 2 8 2 · 3! 16 13. With r = −1/2,
sin
−1
x
x= 0
x
= 0
(1 − t2 )(−1/2) dt ⎡ 1 ⎢ 2 ⎣1 − (−t ) + 2
− 12
− 12 2!
−1
(−t2 )2 +
−
⎤
1 1 − 2 − 1 − 12 − 2 ⎥ 2 (−t2 )3 + · · · ⎦ dt 2!
3 4 3·5 1 t + 3 + · · · dt 1 + t2 + 2 2 2 · 2! 2 · 3! 0 1 3 3 3·5 x + 2 x5 + 3 x7 + · · · =x+ 2·3 2 · 2! · 5 2 · 3! · 7 ∞ 1 · 3 · 5 · · · (2k − 1) 2k+1 =x+ x . 2k k!(2k + 1)
x
=
k=1
bx b√ 2 a − x2 . Then y = − √ a a a 2 − x2 * a* 1 a a4 − x2 (a2 − b2 ) b2 x 2 dx = 1+ 2 2 dx. L= a (a − x2 ) a 0 a2 − x2 0
14. (a) The equation of the ellipse can be written as y = and
Let k 2 a2 = a2 − b2 and x = a sin θ. The dx = a cos θdθ and * * * a π/2 a2 − k 2 x2 a2 − k 2 a2 sin2 θ 1 a a4 − a2 k 2 x2 a cos θdθ L= dx = dx = a 0 a2 − x2 a2 − x2 a2 − a2 sin2 θ 0 0 * π/2 π/2 1 − k 2 sin2 θ =a 1 − k 2 sin2 θdθ. cos θdθ = a 2 1 − sin θ 0 0
9.11. BINOMIAL SERIES
659
(b) With r = 1/2,
π/2
a 0
−1 2 2 2 (−k sin θ) + · · · dθ 2! 0 π/2 1 1 =a 1 − k 2 sin2 θ − k 4 sin4 θ − · · · dθ 2 8 0 1 1 1 θ − sin 2θ = a θ − k2 2 2 4 π/2 1 4 3 1 1 − k θ − sin 2θ + sin 4θ − · · · 8 8 4 32 0 π 1 2 π 1 4 3π − k − k − ··· =a 2 2 4 8 16 π a π a 3π 4 k − ··· . =a − · − · 2 2 4 8 16
1 − k 2 sin2 θdθ = a
π/2
1 1 + (−k 2 sin2 θ) + 2
1 2
1 2
8d x. Using the formula for arc length and r = 1/2, we have l2
2 l/2 l/2 64d2 2 64d2 2 1 64d2 2 12 12 − 1 1 + 4 x dx = 2 x + · · · dx s=2 1+ · 4 x + l 2 l 2! l4 0 0 l/2 l/2 4 32d2 3 642 d4 5 32d2 2 642 d4 + · · · dx = 2 x + x − x + · · · 1+ 4 x − =2 8 4 8 l 8l 3l 5 · 8l 0 0
15. y =
=l+
0.2
16. (a) 0
32d4 8d2 − + ··· . 3l 5l3
− 1 3 2 (x ) + · · · dx 1 + x3 dx = 2! 0 0.2 0.2 1 1 1 1 1 + x3 − x6 + · · · dx = x + x4 − x7 + · · · = 2 8 8 56 0 0
0.2
= 0.2 +
1 1 + x3 + 2
1 2
1 2
(0.2)7 (0.2)4 − + ··· 8 56
(0.2)7 After the first term this is an alternating series. Since < 0.0005, we have will 56 4 0.2 √ (0.2) three decimal place accuracy 0 = 0.2002. 1 + x3 dx ≈ 0.2 + 8
1/2 1/2 1 1 −1 1 3 (x4 )2 + · · · dx 1 + x4 dx = (b) 1 + x4 + 3 3 3 2! 0 0 1/2 1/2 2 8 1 5 2 1 4 9 = x + · · · 1 + x − x + · · · dx = x + x − 3 18 15 18 · 9 0 0 1 2 1 − = + + ··· 2 15 · 32 18 · 9 · 29
CHAPTER 9. SEQUENCES AND SERIES
660
2 After the first term this is an alternating series. Since < 0.0005, we have with 18 · 9 · 29 1/2 √ 1 1 ≈ 0.502. three decimal place accuracy 0 3 x + x4 ≈ + 2 15 · 32 17. From Theorem 4.11.1 9.11.1 in the text with r = −1/2, we have
− 12 − 12 − 1 2 1 2 −1/2 −1/2 (1 − 2xr + r ) r (r − 2x)2 + · · · = [1 + r(r − 2x)] = 1 − r(r − 2x) + 2 2! 3 2 1 3 1 x − r2 + · · · = 1 − r2 + xr + (r4 − 4xr3 + 4x2 r2 ) + · · · = 1 + xr + 2 8 2 2 3 1 Thus, P0 = 1 P1 = x, and P2 = x2 − . 2 2 r(r − 1)(r − 2) 2 r(r − 1) · · · (r − n + 1) n−1 x + ··· + x + ··· 2! (n − 1)! r(r − 1)(r − 2) 3 r(r − 1) · · · (r − n + 1) n x + ··· + x + ··· xf (x) = rx + r(r − 1)x2 + 2! (n − 1)! r(r − 1) · · · (r − n + 1) r(r − 1) · · · (r − n) +n (b) (n + 1) (n + 1)! n! r(r − 1) · · · (r − n) nr(r − 1) · · · (r − n + 1) + = n! n! r(r − 1) · · · (r − n + 1) r(r − 1) · · · (r − n + 1) (r − n + n) = r = n! n! r(r − 1) · · · (r − n + 1) xn−1 + · · · (c) f (x) + xf (x) = r + r(r − 1)x + · · · + (n − 1)! r(r − 1) · · · (r − n + 1) n rx + r(r − 1)x2 + · · · + x + ··· (n − 1) ∞ ∞ r(r − 1) · · · (r − k + 1) k−1 r(r − 1) · · · (r − j + 1) j x x + = (k − 1)! (j − 1)! j=1
18. (a) f (x) = r + r(r − 1)x +
k=1
Let k = j + 1 ∞
∞ r(r − 1) · · · (r − j) j r(r − 1) · · · (r − j + 1) j x + x = j! (j − 1)! j=0 j=1 ∞ r(r − 1) · · · (r − j) r(r − 1) · · · (r − j + 1) j + =r+ x j! (j − 1)! j=1 ∞ r(r − 1) · · · (r − j + 1) j r(r − 1) · · · (r − j) +j =r+ (j + 1) x (j + 1)! j! j=1
By part (b) ∞ r(r − 1) · · · (r − j + 1) j x =r+ r j! j=1 r(r − 1) 2 r(r − 1) · · · (r − n + 1) n x + ··· x + · · · = rf (x) =r 1+r+ 2 n!
CHAPTER 9 IN REVIEW
661
(d) Write the equation in the form (1 + x)
dy = ry. Then, separating variables, we have dx
r dy rdx = ; ln y = r ln(1 + x) + c; y = er ln(1+x)+c = ec eln(1+x) = c1 (1 + x)r y 1+x
or f (x) = c1 (1 + x)r . Now, 1 + f (0) = c1 so f (x) = (1 + x)r . 1/2 √ x−1 1/2 1/2 19. (1 + x) = [2 + (x − 1)] = 2 1+ 2
1 1
1 1 1 2 4 √ 1 x−1 2 2 − 1 (x − 1) 2 2 −1 2 − 2 (x − 1) + = 2 1+ + + ··· 2 2 2! 22 3! 24 √ √ √ √ 2 2 2·1·3 (x − 1)3 + · · · = 2 + 2 (x − 1) − 2 (x − 1)2 + 2 2 2! 26 · 3! −2 x−1 −2 −2 −2 20. (1 + x) = [2 + (x + 1)] = 2 1+ 2 1 −2(−2 − 1)(−2 − 2) (x − 1)3 x − 1 −2(−2 − 1) (x − 1)2 + = + 1−2 4 2 2! 22 3! 23 3 3·4 1 1 (x − 1)2 − 4 (x − 1)3 + · · · = − (x − 1) + 2 4 4 2 · 2! 2 3!
PROBLEMAS Chapter 9 DE in REPASO ReviewDE LA UNIDAD 4 A. True/False 1. False; since |an | =
1 n → , the series diverges by the n-th term test. 2n + 1 2
2. False; {(−1)n } is bounded and divergent. 3. False; {(−1)n /n} is convergent and not monotonic. 4. True; the first three terms are 1/2, 100/6, and 1000/512. 5. True; since an+1 /an ≥ 1, an+1 ≥ an and {an } is a bounded monotonic sequence. 6. True 7. False; {1/n} converges, but
∞
1/k is the divergent harmonic series.
k=1
9 1 1 1 + + + ··· ) 1+ 10 10 100 1000 9 1 1 9 = = =1 1 9 10 1 − 10 10 10
8. True; 0.999 . . . =
9. True; if this were false, the series would diverge by the n-th term test.
CHAPTER 9. SEQUENCES AND SERIES
662 10. False; consider the harmonic series 11. False; let ak = 1/k. 12. True 13. True 14. True 15. False; let ak = (−1)k /k. 16. True 17. True 18. False; let bk = 1/k 2 and ak = a/k. 19. False; the ratio test is inconclusive in this case. 20. False;
∞
k!xk converges only at x = 0.
k=0 ∞ 1 k x converges at x = −1 but is not absolutely convergent there. 21. False; k k=1
22. True 23. False; let ck = −1/k. 24. False; at x = r the series diverges 25. False; the integral test simply indicates that the series converges. 26. True; the series is absolutely convergent 27. True; ln x is not differentiable at x = 0. 28. True 29. True 30. True; f (4) = 4!c4 = − since c4 = 0.
B. Fill in the Blanks 1. 20, 9, 4/5, 16 2. π/2 3. 4, since a5 = 1/105 = 0.00001 < 0.00005. 4. 12
CHAPTER 9 IN REVIEW 5.
n 22 ; 9 9
663
1 1 + 2 + ... 1+ 10 10 ⎛ ⎞ n ⎜ 1 ⎟ 1 n n = = = ⎝ 1 ⎠ 10 9 10 9 10 1− 10 4 22 2.444 . . . = 2 + .044 . . . = 2 + = 9 9 0.nnnn . . . =
n 10
6. −π/4 7. ex 8. 4 9. x < −5 and x > 5. 10.
∞ (−x3 )k k=0
k!
=
∞ (−1)k x3k k=0
k!
11. The series converges absolutely for x in (−1, 1). At x = −1, the series diverges. At x = 1, the series converges conditionally. Thus, the interval of convergence is (−1, 1]. 12. 5
C. Exercises ∞ 1 k 1 1. Since 2 < 3 , the series converges by comparison with the p-series . (k + 1)2 k kp k=1
2. Since lim
n→∞
1 = 1, the series diverges by the n-th term test. 1 + e−k
3. This is a geometric series with r =
1 < 1. Thus, the series converges. π
4. This is a geometric series with r =
1 > 1. Thus, the series diverges. ln(2.5)
√ 5. Since ln k < k, ∞ 1 5/2 k k=1
1 k ln k < 5/2 and the series converges by comparison with the p-series k4 + 4 k
sin k 1 6. Since 3/2 < 3/2 , the series is absolutely convergent, and thus convergent, by comparison k k ∞ 1 with the p-series . 3/2 k k=1
CHAPTER 9. SEQUENCES AND SERIES
664
1 k k > 2 = , the series diverges by comparison with the harmonic series. k k − 4k √ 8. The function f (x) = 1/x ln x is continuous and decreasing on [2, ∞). Since
7. Since √ 3
k6
∞ 2
dx √ = lim x ln x t→∞
t 2
dx √ x ln x
ln t
= lim
t→∞
ln 2
u = ln x, du =
1 dx x
√ √ √ ln t du √ = lim 2 uln 2 = lim 2 ln t − ln 2 = ∞, t→∞ u t→∞
the series diverges by the integral test. 9. All of the odd-numbered terms of this series are 0. We may thus express the series as or
∞ 1 √ . This is a divergent p-series. k k=1
∞ 2 √ 2k k=1
10. Since 2 an+1 [(n + 1)2 !]/[(n + 1)!] (n2 + 2n + 1)! n! = lim = lim n→∞ an n→∞ n→∞ (n2 )!/(n!)2 (n2 )! (n + 1)! 2 1 = lim (n2 + 2n + 1)(n2 + 2n) · · · (n2 + 1) n→∞ n+1 (n2 + 2n + 1)(n2 + 2n) · · · (n2 + 1) = lim n→∞ (n2 + 2n + 1) lim
= lim (n2 + 2n) · · · (n2 + 1) = ∞, n→∞
the series diverges by the ratio test. ∞ 1 1 1 < , the series converges by comparison with the p-series . 2 2 3k + 4k + 6 k k2 k=1 3n 12. an = ln → ln(3) = 0. Thus, the series diverges by the n-th term test. n+1 k−1 k−1 ∞ ∞ ∞ 1 3 (−1)k−1 + 3 1 1 + = +3 = − 13. (1.01)k−1 1.01 1.01 1 − (−1/1.01) 1 − (1/1.01)
11. Since
k=1
k=1
k=1
1.01 3.03 101 61, 004 = + = + 303 = 1.01 + 1 1.01 − 1 201 201 1 1 1 1 1 1 1 1 − . Then Sn = − − − 14. Write ak = + + ··· + = k+1 k+6 6 7 7 8 n+5 n+6 1 1 − 6 n+6 n 1 1 = lim Sn = . and lim 2 n→∞ n→∞ k + 11k + 30 6 k=1
CHAPTER 9 IN REVIEW
665
n+1 n+1 3 an+1 n x /(n + 1)3 lim 3 = lim 15. lim 3 |x| = 3|x| n→∞ n→∞ an n→∞ 3n xn /n3 n+1 The series is absolutely convergent for 3|x| < 1 or on (−1/3, 1/3). At x = −1/3, the se∞ ∞ (−1)k 1 ries converges by the alternating series test. At x = 1/3, the series is a 3 k k3 k=1 k=1 convergent p-series. Thus, the given series converges on [−1/3, 1/3]. an+1 (n + 1)(2x − 1)n+1 /4n+1 1 n+1 1 lim lim 16. lim |2x − 1| = |2x − 1| = n→∞ n→∞ an n→∞ n(2x − 1)n /4n 4 n 4 1 The series is absolutely convergent for |2x − 1| < 1, |2x − 1| < 4 or on (−3/2, 5/2). At 4 ∞ ∞ k x = −3/2, the series (−1) k diverges by the n-th term test. At x = 5/2, the series k k=1
diverges by the n-th term test. Thus, the given series converges on (−3/2, 5/2).
k=1
n+1 an+1 lim (n + 1)!(x + 5) = lim (n + 1)|x + 5| = ∞ for x = −5. Thus, the series 17. lim n→∞ n→∞ an n→∞ n!(x + 5)n converges only for x = −5. an+1 (2x)n+1 / ln(n + 1) 18. lim = lim n→∞ an n→∞ (2x)n / ln n ln n 1/n h |2x| = lim |2x| = lim n→∞ ln(n + 1) n→∞ 1/(n + 1) n+1 = lim |2x| = |2x| n→∞ n The series is absolutely convergent for |2x| < 1 or on (−1/2, 1/2). At x = −1/2, the series ∞ 1 diverges by comparison with the harmonic series. Thus, the given series converges ln 2 k=2 on [−1/2, 1/2). 2 · 5 · · · (3n + 2)xn+1 an+1 = lim 3 · 7 · · · (4n + 3) = lim 3n + 2 |x| = 3 |x 19. lim n n→∞ 4n + 3 n→∞ n→∞ 2 · 5 · · · (3n − 1)x an 4 3 · 7 · · · (4n − 1) 3 4 The series converges for |x| < 1 or |x| < . Thus, the radius of convergence is 4/3. 4 3 an+1 (cos x)n+1 = lim | cos x| = | cos x|. = lim 20. Applying the ratio test, we have lim n→∞ an n→∞ (cos x)n n→∞ Since | cos x| < 1 for x = kπ where k is an integer, the series converges for all x = kπ. When ∞ ∞ x = kπ, the series is either (−1)k or 1, both of which diverge by the n-th term test. k=1
k=1
Thus, the series converges on all intervals of the form (kπ, kπ + π), where k is an integer.
CHAPTER 9. SEQUENCES AND SERIES
666
1 1 1 + + 2 + ··· − 1 α α 1 1 α
−1= −1= = α−1 α−1 1 − α1
1 1 1 + 2 + 3 + ··· = 21. α α α
22. The argument is invalid since both series diverge. This implies S = ∞ and hence S cannot be subtracted from both sides of 2S = S − 1. 23. Using a binomial series expansion,
1 1 −3 −3 − 1 1 5 1 2 1 5 −1/3 √ (x5 )2 + · · · = 1 − x5 + x10 − · · · . = (1 + x ) =1− x + 3 3 2! 3 9 1 + x5 24. Using a binomial series expansion, x 2−x−2 2 x −1 = = −1 + = −1 + 1 − 2−x 2−x 2−x 2 x (−1)(−1) x 2 (−1)(−1 − 1)(−1 − 2) x 3 = −1 + 1 + (−1) − + + ··· + − − 2 2! 2 3! 2 1 2 1 3 1 = x + x + x + ··· . 2 4 8 25. Using the Maclaurin series for sin x, 1 1 (2x)5 2 2 (2x)3 sin x cos x = sin 2x = + − · · · = x − x3 + x5 − · · · . 2x − 2 2 3! 5! 3 15 26. Using the Maclaurin series for et , x x x 1 2 2 1 4 t2 2 2 e dt = 1 + t + (t ) + · · · dt = 1 + t + t + · · · dt 2 2 0 0 0 x 1 1 1 1 = 1 + t3 + t5 + · · · = x + x3 + x5 + · · · . 3 10 3 10 0 27. f (x) = cos x,
f (π/2) = 0
f (x) = − sin x,
f (π/2) = −1
f (x) = − cos x,
f (π/2) = 0
f (x) = sin x, .. .
f (π/2) = 1
f (2k+1) (π/2) = (−1)k+1 f (2k+1) (x) = (−1)k+1 sin x, ∞ (−1)k+1 π 2k+1 x− The Taylor series is . (2k + 1)! 2 k=0
CHAPTER 9 IN REVIEW
667
28. We use |f (n+1) (x)| =
2n−1 | sin(2x)|, n−1
2
| cos(2x)|,
|Rn (x)| =
n even n odd
Since | cos(2x)| ≤ 1 and | sin(2x)| ≤ 1,
|f (n+1) (c)| n+1 2n |x|n+1 |x| ≤ (n + 1)! (n + 1)!
|an+1 | 2n+1 |x|n+2 (n + 1)! = n n+1 · an 2 |x| (n + 2)! 2|x| . = n+2 which converges to 0 as n → ∞. Thus, the series found in Problem 55 represents sin x cos x for all x. 2 2 4 2 8 2 16 2 29. 3 +2 + + + + ··· 3 3 3 3 9 3 27 3 3 4 5 2 2 2 2 2 2 2 = 6 million dollars. +3 +3 +3 + ··· = =3 +3 3 3 3 3 3 1 − 2/3 Using the ratio test,
30. (a) Solving 2P = P (1 + r)n for n, we obtain the doubling time n = ln 2/ ln(1 + r). (b) Since 0 < r < 1, ln(1 + r) = r − r2 /2 + r3 /3 − · · · ≈ r. Then (ln 2)/ ln(1 + r) ≈ (ln 2)/r ≈ 0.69/r ≈ 70/100r. (c) We want to solve 70/100r = ln 2/ ln(1+r) or 7 ln(1+r) = 10r ln 2 for r. Using ln(1+r) ≈ r − r2 /2 + r3 /3, this equation can be written as 7 2 7 7 3 7 2 r − r + (7 − 10 ln 2)r = r r − r + 7 − 10 ln 2 = 0. 3 2 3 2 The quadratic formula gives r ≈ 1.4802 and r ≈ 0.0198. Since r < 1 we see that the Rule of 70 gives the true doubling time for r ≈ 1.98%.
73. Rewrite the integral as
x5 dx = (x − 1)10 (x + 1)10
x5 dx = (x2 − 1)10
u = x2 − 1, x2 = u + 1,
then integrate using
4
x4 (x dx) (x2 − 1)10
1 u = x dx : 2
CÁLCULO x (u + 1) u 1 INTEGRAL 1 (x dx) = du = (x − 1) 2 u 2 MATEMÁTICAS 2 1 2
2
2
+ 2u + 1 du u10 1 1 1 (u−8 + 2u−9 + u−10 ) du = − u−7 − u−8 − u−9 + C = 2 14 8 18 1 2 1 2 1 2 −7 −8 −9 = − (x − 1) − (x − 1) − (x − 1) + C 14 8 18
10
10
APÉNDICE INTEGRALES IMPROPIAS 74. The integrand in Problem 53 is an odd function and its definite integral is symmetric about the y-axis. Thus, the definite integral’s value is known to be 0.
7.7
Improper Integrals h
In this exercise set, the symbol “=” is used to denote the fact that L’Hˆ opital’s Rule was applied to obtain the equality.
∞
1. 3
−1
2. −∞
1 dx = lim t→∞ x4
t
x
−4
dx = lim
t→∞
3
1 √ dx = lim 3 s→−∞ x
−1
x
−1/3
s
1 − x−3 3
t
= lim
t→∞
3
3 2/3 x dx = lim s→−∞ 2
1 1 − 81 3t3
−1
= lim
s
s→−∞
=
3 3 2/3 − s 2 2
The integral diverges.
∞
3. 1
1 x0.99
t
dx = lim
t→∞
The integral diverges.
x 1
−0.99
x0.01 dx = lim t→∞ 0.01
t = lim 1
t→∞
1 81
t0.01 1 − 0.01 0.01
7.7. IMPROPER INTEGRALS
481
t 1 x−0.01 1 − = lim = 100 t→∞ 0.01 t→∞ 1 t→∞ −0.01 x1.01 0.01t0.01 1 1 3 3 3 1 6 1 2x 1 2x 1 2x 2x e e − e e dx = lim e dx = lim = lim = e6 5. s→−∞ s→−∞ s→−∞ 2 2 2 2 −∞ s s ∞ 0 ∞ 6. e−x dx = e−x dx + e−x dx
∞
4.
−∞
1
dx = lim
−∞
e−x dx = lim
s→−∞
the integral diverges. ∞ t ln x 7. dx = lim t→∞ 1 x 1 The integral diverges. ∞ k ln 2 8. dt = lim k→∞ 1 t2 1
10.
11.
12.
x−1.01 dx = lim
−∞ 0
Since
9.
t
0
0 s
e−x dx = lim (−e−x ) s→−∞
ln x 1 dx = lim (ln x)2 t→∞ x 2
0 s
t
s→−∞
= lim
1
= lim (e−s − 1),
t→∞
1 (ln t)2 − 0 2
1 ln t 1 1 dt u = ln t, du = dt; dv = 2 dt, v = − t2 t t t k k k 1 1 1 1 = lim − ln t + dt = lim 0 − ln k − 2 k→∞ k→∞ t t k t 1 1 1 1 ln k h 1/k 1 =1 = 1 − lim = lim 1 − − ln k = 1 − lim k→∞ k→∞ k k→∞ 1 k k t ∞ t
1 1 1 1 1 −3 1 −2
dx = lim (ln x) − dx = lim (ln x) = lim − =
3 2 t→∞ t→∞ t→∞ x(ln x) x 2 2 2(ln t) 2 e e e ∞ t t ln t − 1 ln x dx = lim ln x dx = lim (x ln x − x) = lim (t ln t − t − e + e) = lim t→∞ t→∞ t→∞ t→∞ 1/t e e e The limit has the form ∞/0, so the integral diverges. 0 ∞ ∞ x x x dx = dx + dx 2 3/2 2 3/2 2 (x + 1)3/2 −∞ (x + 1) −∞ (x + 1) 0 0 t x x dx + lim dx = lim s→−∞ s (x2 + 1)3/2 t→∞ 0 (x2 + 1)3/2 0 t = lim [−(x2 + 1)−1/2 ] + lim [−(x2 + 1)−1/2 ] s→−∞ t→∞ s 0 1 1 + lim − √ + 1 = −1 + 1 = 0 = lim −1 + √ s→−∞ t→∞ s2 + 1 t2 + 1 0 ∞ ∞ x x x dx = dx + dx 2 2 1 + x 1 + x 1 + x2 −∞ −∞ 0 0 0 0 x x 1 2 ln(1 + x Since dx = lim dx = lim ) 2 s→−∞ s 1 + x2 s→−∞ 2 −∞ 1 + x s 1 1 2 [ln 1 − ln(1 + s )] = − lim ln(1 + s2 ), = lim s→−∞ 2 2 s→−∞
CHAPTER 7. TECHNIQUES OF INTEGRATION
482 the integral diverges.
0
13. −∞
∞
14. 5
0
x 1 2 −1
(x dx = lim + 9) −
2 2 s→−∞ 2 s (x + 9) s 1 1 1 1 = lim − =− s→−∞ 2 s2 + 9 18 18
x dx = lim s→−∞ (x2 + 9)2
1 √ dx = lim 4 t→∞ 3x + 1
0
t
4 (3x + 1) dx = lim (3x + 1)3/4 t→∞ 9 5 4 4 (3t + 1)3/4 − (8) = lim t→∞ 9 9 −1/4
t 5
The integral diverges.
∞
15.
ue
−u
t
du = lim
t→∞
2
2
t
ue−u du = lim (−ue−u − e−u )
2
t→∞
= lim (2e−2 + e−2 − te−t − e−t ) t→∞
t + 1 h −2 1 = 2e + e−2 − lim t = 2e−2 + e−2 = 3e−2 t→∞ e et 3 3 x3 x3 1 4 dx = lim dx = lim ln(x + 1) s→−∞ s x4 + 1 s→−∞ 4 x4 + 1 s 1 1 ln 82 − ln(s4 + 1) = lim s→−∞ 4 4 = 2e−2 + e−2 − lim
t→∞
3
16. −∞
The integral diverges.
∞
17. 2/π
sin(1/x) dx = lim t→∞ x2
∞
te
18.
−t2
0
dt =
te −∞
−∞
−t2
t 2/π
sin(1/x) 1 dx = lim cos 2 t→∞ x x
∞
dt +
te 0
0
−t2
= lim 2/π 0
dt = lim
s→−∞
t
te s
r
−t2
t→∞
cos
π 1 − cos t 2
r
dt + lim
r→∞
=1
2
te−t dt
0
2 2 1 1 + lim − e−t − e−t r→∞ 2 2 s 0 1 −s2 1 1 1 −r2 1 1 e − e = lim − =− + =0 + lim s→−∞ 2 r→∞ 2 2 2 2 2
=
∞
19. −1
0
20. −∞
lim
s→−∞
t 1 −1 dx = lim tan (x + 1) 2 t→∞ −1 −1 (x + 1) + 1 π = lim [tan−1 (t + 1) − 0] = t→∞ 2 0 0 1 1 1 −1 x + 1 √ dx = lim dx = lim √ tan s→−∞ s (x + 1)2 + 2 s→−∞ x2 + 2x + 3 2 2 s 1 1 s + 1 1 1 1 π √ tan−1 √ − √ tan−1 √ = √ tan−1 √ + √ = lim s→−∞ 2 2 2 2 2 2 2 2
1 dx = lim t→∞ x2 + 2x + 2
t
7.7. IMPROPER INTEGRALS
∞
21. 0
e−x sin x dx = lim
t→∞
t
483 e−x sin x dx
0
u = e−x , du = −e−x dx; dv = sin x dx, v = − cos x t t −x −x = lim −e cos x − e cos x dx 0
t→∞
−x
0
−x
u = e , du = −e dx; dv = cos x dx, v = sin x t t −t −x −x = lim 1 − e cos t − e sin x − e sin x dx t→∞ 0 0 t ∞ = lim 1 − e−t cos t − e−t sin t − e−x sin x dx = 1 − e−x sin x dx t→∞ 0 0 ∞ 1 Solving for the integral, e−x sin x dx = . 2 0 0 0 x 22. e cos 2x dx = lim ex cos 2x dx −∞
s→−∞
s
1 u = ex , du = ex dx; dv = cos 2x dx, v = sin 2x 2 0 0 1 x 1 e sin 2x − = lim ex sin 2x dx s→−∞ 2 2 s s 1 u = ex , du = ex dx; dv = sin 2x dx, v = − cos 2x 2 0 0 1 1 1 lim =− ex cos 2x dx − ex cos 2x + 2 s→−∞ 2 2 s s 0 1 1 1 =− ex cos 2x dx − −0 − 2 2 4 −∞ 0 4 1 1 x Solving for the integral, e cos 2x dx = = . 5 4 5 −∞ t ∞ t x+1 1 −2 −2 −3 −1 (x + x ) dx = lim − −x x 23. dx = lim 3 t→∞ 1/2 t→∞ 2 1/2 x 1/2 1 1 = lim 2 + 2 − − 2 = 4 t→∞ t 2t t ∞ (e−x − e−2x )2 dx = lim (e−2x − 2e−3x + e−4x ) dx 24. 0
t→∞
0
t 1 −2x 2 −3x 1 −4x = lim − e + e − e t→∞ 2 3 4 0 1 −2t 2 −3t 1 −4t 1 2 1 1 = lim − e + e − e + − + = t→∞ 2 3 4 2 3 4 12
CHAPTER 7. TECHNIQUES OF INTEGRATION
484
∞
25. 1
26.
27.
28.
29.
30.
31. 32. 33.
1 1 − x x+1
t
dx = lim
t→∞
1
1 1 − x x+1
dx = lim (ln |x| − ln |x + 1|) t→∞
t 1
t = lim [ln t − ln(t + 1) + ln 2] = ln 2 + lim ln t→∞ t→∞ t+1 t 1 h = ln 2 + ln lim = ln 2 + ln lim = ln 2 + ln 1 = ln 2 t→∞ t + 1 t→∞ 1 t t ∞ 1 1 1 1 1 −1 x + + dx = lim dx = lim ln |x| + tan t→∞ 3 t→∞ x x2 + 9 x x2 + 9 3 3 3 3 π t 1 = lim ln t + tan−1 − ln 3 − t→∞ 3 3 12 The integral diverges. t t ∞ 1/4 1/4 1 1 dx = lim dx = lim − dx t→∞ 2 (x + 1)(x + 5) t→∞ 2 x2 + 6x + 5 x+1 x+5 2
t t 1 1 1 x + 1
ln |x + 1| − ln |x + 5| = lim ln
= lim t→∞ 4 t→∞ 4 4 x + 5 2 2 1 t+1 1 3 t+1 1 1 3 ln − ln = lim = ln lim − ln t→∞ 4 t→∞ t + 5 t+5 4 7 4 4 7 3 1 3 1 7 1 1 1 h = ln lim − ln = − ln = ln t→∞ 1 4 4 7 4 7 4 3 0 0 0 1 1 1 dx = − dx = lim (ln |x − 2| − ln |x − 1|) 2 s→−∞ x−2 x−1 s −∞ x − 3x + 2 −∞
0
x − 2
= ln 2 − ln 1 = ln 2 = lim ln
s→−∞ x − 1 s −2 −2 −2
x2 1 3x2 1 1
dx = lim dx = lim −
3 + 1)2 3 + 1)2 3+1 s→−∞ s→−∞ (x 3 (x 3 x −∞ s s 1 1 1 =− − = 3 7 21 t et ∞ 1 ex 1 x x dx du dx = lim u = e , du = e dx = lim x + e−x 2x + 1 2+1 t→∞ t→∞ e e u 1 0 0 et π π π π = lim tan−1 et − = lim tan−1 u = − = t→∞ t→∞ 4 2 4 4 1 5 5 5 1 1 dx = lim dx = lim ln |x| = lim (ln 5 − ln s). The integral diverges. s s→0+ s x s→0+ s→0+ 0 x 8 8 8 1 −2/3 1/3 dx = lim x dx = lim 3x = lim+ (6 − 3s1/3 ) = 6 2/3 s s→0+ s s→0+ s→0 0 x 1 1 1 1 −0.99 0.01 dx = lim x dx = lim 100x = lim+ (100 − 100s0.01 ) = 100 0.99 s s→0+ s s→0+ s→0 0 x
7.7. IMPROPER INTEGRALS
1
34.
1 x1.01
0
dx = lim+ s→0
1
485
x−1.01 dx = lim+ (−100x−0.01 ) s→0
s
1 s
= lim+ s→0
100 − 100 s0.01
The integral diverges. 2 t t 1 √ 35. (2 − x)−1/2 dx = lim− [−2(2 − x)1/2 ] dx = lim− 0 t→2 t→2 2−x 0 0 √ √ √ = lim− (2 2 − 2 2 − t) = 2 2 t→2
3
36. 1
1 dx = lim+ (x − 1)2 s→1
3
(x − 1)
−2
dx = lim+ [−(x − 1)
−1
]
3 s
s→1
s
= lim+ s→1
1 1 − s−1 2
The integral diverges. t 1 1 1 1 1 dx = lim dx + lim dx 37. 5/3 5/3 5/3 − + t→0 s→0 −1 x −1 x s x t t 3 1 3 3 −2/3 − dx = lim− − x = lim− Since lim− , the integral diverges. 5/3 2 2 2t2/3 t→0 t→0 t→0 −1 x −1
2
√ 3
38. 0
1 dx = lim− t→1 x−1
t 0
(x − 1)−1/3 dx + lim+
s→1
2
(x − 1)1/3 dx
s
t 2 3 3 (x − 1)2/3 + lim (x − 1)2/3 t→1 2 s→1+ 2 0 s 3 3 3 3 3 3 2/3 − (s − 1)2/3 = − + = 0 = lim− (t − 1) − + lim+ 2 2 2 2 2 2 t→1 s→1 t 2 2 (x − 1)−2/3 dx = lim− (x − 1)−2/3 dx + lim+ (x − 1)−2/3 dx 39. = lim−
t→1
0
0
= lim− 3(x − 1)1/3 t→1
27
40. 0
1/3
ex dx = lim+ s→0 x2/3
= lim+ s→0
1
0
x−2/3 ex
1/3
0
dx
s→1
+ lim+ 3(x − 1)1/3
s1/3
u = x1/3 , du =
3eu du = lim+ 3eu
1
s→0
2
s→1
s 3
s
3 s1/3
=3+3=6
s
1 −2/3 x dx 3
1/3 = lim+ 3e3 − 3es = 3e3 − 3 s→0
1 1 u = ln x, du = dx; dv = x dx, v = x2 x 2 s 1 1 1 1 1 1 2 1 x ln x − x dx = lim − s2 ln x − x2 = lim + + 2 2 4 s→0 s→0 s 2 s s 2 1 s 1 ln s 1 = lim − s2 ln s − + − = lim − 2 4 4 2/s2 4 s→0+ s→0+ 1 s2 1 1 1/s h = lim − =− = lim − − 3 + + −4/s 4 4 4 4 s→0 s→0
x ln x dx = lim
41.
27
t
s→0+
x ln x dx
CHAPTER 7. TECHNIQUES OF INTEGRATION
486
e
42. 1
1 dx = lim x ln x s→1+
e
1 ln x
s
1 dx x
= lim ln(ln x) +
e
s→1
s
= lim+ [0 − ln(ln s)] s→1
The integral diverges.
π/2
43.
tan t dt = 0
k
lim
k→π/2−
tan t dt = 0
lim
k→π/2−
ln sec t
k 0
=
lim
k→π/2−
ln sec k
The integral diverges.
π/4
44. 0
sec2 θ √ dθ = lim s→0+ tan θ
π/4 s
1
= lim
s→0+
π
45. 0
sin x dx = lim 1 + cos x t→π −
tan s
= lim− t→π
t 0
sec2 θ √ dθ u = tan θ, du = sec2 θ dθ tan θ √ √ 1 1 √ du = lim 2 u = lim+ (2 − 2 tan s) = 2 u tan s s→0+ s→0
sin x dx 1 + cos x
1+cos t
2
u = 1 + cos x, du = − sin x dx
1+cos t 1 − du = lim− (− ln |u|) = lim− [ln 2 − ln(1 + cos t)] u 2 t→π t→π
The integral diverges. π t π cos x cos x cos x √ √ √ 46. dx = lim dx + lim dx t→π/2− 0 s→π/2+ s 1 − sin x 1 − sin x 1 − sin x 0 t π √ √ = lim −2 1 − sin x + lim −2 1 − sin x = 2 + (−2) = 0 t→π/2−
0
47. −1
√
x dx = lim s→−1+ 1+x
0 s
0
√
x dx 1+x
s→π/2+
s
u2 = 1 + x, x = u2 − 1, dx = 2u du
1 u2 − 1 (2u du) = lim (2u2 − 2) du √ √ + u s→−1 s→−1 1+s 1+s 1 √ 2 3 4 4 2 3/2 u − 2u √ = lim + − − (1 + s) + 2 1 + s = − = lim + 3 3 3 3 s→−1 s→−1 1+s 1
= lim +
48.
t 3 1 1 1 dx = lim dx + lim dx 2−1 2−1 2−1 − + x x x t→1 s→1 0 0 s Since t t t 1/2 1 1/2 1 1 dx = lim − ln |x − 1| − ln |x + 1| lim dx = lim x−1 x+1 2 2 t→1− 0 x2 − 1 t→1− 0 t→1− 0
t
x − 1
t − 1
1 1 , = lim− ln
= lim− ln
x + 1
t + 1
t→1 2 t→1 2 3
0
the integral diverges.
7.7. IMPROPER INTEGRALS
1
49. 0
x2 √ dx = lim− t→1 1 − x2
= lim− t→1
t 0
487
1
x2 √ dx 1 − x2
sin
−1
t
0
x = sin θ, dx = cos θ dθ
sin2 θ 1 − sin2 θ
cos θ dθ = lim− t→1
sin
−1
x
θ
√ t
1 − x2
2
sin θ dθ 0
sin−1 t 1 1 1 (1 − cos 2θ) dθ = lim θ − sin 2θ = lim− 2 2 4 t→1 t→1− 0 0 sin−1 t
1 1 1 1 π θ − sin θ cos θ sin−1 t − t 1 − t2 = = lim = lim − − 2 2 2 2 4 t→1 t→1 0
2
50. 0
√
ew dw = lim s→0+ ew − 1
sin−1 t
2
√
s
ew dw ew − 1
√ √ e −1 1 2 − 1 − 2 es − 1 √ du = lim 2 u = lim+ = lim e 2 u es −1 s→0 s→0+ s→0+ es −1
= 2 e2 − 1
3
51. 1
e2 −1
1 √ dx = lim t→3− 3 + 2x − x2
2
t 1
t→3−
1
52. 0
1 1 √ +√ x 1−x
1
= lim sin−1
u = ew − 1, du = ew dw
dx = lim sin t→3− 4 − (x − 1)2 t−1 π = sin−1 1 = 2 2
−1
x−1 2
t 1
t 1 1 1 1 √ +√ √ +√ dx = lim+ dx + lim− dx x x s→0 t→1 1−x 1−x s 1/2 1/2 t √ √ √ √ + lim− (2 x − 2 1 − x) = lim+ (2 x − 2 1 − x) s 1/2 s→0 t→1 √ √ √ √ = lim+ ( 2 − 2 − 2 s + 2 1 − s) s→0 √ √ √ √ + lim− (2 t − 2 1 − t − 2 + 2) = 2 + 2 = 4
1/2
t→1
∞
53. 12
1 √ dx = lim t→∞ x(x + 4)
t 12
√
1 dx x(x + 4)
u=
√
x, u2 = x, dx = 2u du
t u 1 du = 2 lim √ 2 du = 2 lim √ t→∞ 2 3 u(u2 + 4) t→∞ 2 3 u + 4 t √ π 1 u π π −1 t −1 = 2 lim tan−1 − tan = lim 3 = − = tan t→∞ 2 2 2√3 t→∞ 2 2 3 6 t
CHAPTER 7. TECHNIQUES OF INTEGRATION
488
∞
54.
√
xe−
√
x
t
dx = lim
t→∞
1
√
xe−
√
x
dx
w=
√
x, w2 = x, dx = 2w dw
1
t
u = w2 , du = 2w dw; t t = lim 2 −w2 e−w + 2 we−w dw = 2 lim
t→∞
w2 e−w dw
1
1
t→∞
dv = e−w dw, v = −e−w
1
u = w, du = dw; dv = e−w dw, v = −e−w t t t 2 −w −w −w = lim 2 −w e + 2 −we + e dw t→∞ 1 1 1 t t t 2 −w −w −w = lim 2 −w e + 2 −we −e t→∞ 1 1 1 2 −t = lim 2 −t e + e−1 + 2 −te−t + e−1 − e−t + e−1 t→∞
= 2[e−1 + 2(e−1 + e−1 )] = 10e−1 ∞ t 1 1 55. A = dx = lim dx 2 t→∞ (2x + 1) (2x + 1)2 1 1 t 1 1 1/2 1 − = lim − = lim = t→∞ 2x + 1 1 t→∞ 6 4t + 2 6 5 5 10 10 x 5 dx = lim dx = lim 2 tan−1 56. A = 2 2 s→−∞ s x + 25 s→−∞ 5 s −∞ x + 25 π 3π π π s − 2 tan1 = lim = −2 − = s→−∞ 2 5 2 2 2 ∞ ∞ t 57. A = e−|x| dx = 2 e−x dx = 2 lim (−e−x ) −∞
= −2 lim (e
t→∞
0
t→∞
−t
5
s→1
1
-5
5
2
0
1
− e ) = −2(0 − 1) = 2
4
∞
-5
5
4
5 1 2 1 √ √ 59. A = − − dx = lim+ dx s→1 x − 1 x − 1 x −1 1 s 5 √ √ = lim+ 4 x − 1 = lim+ (8 − 4 s − 1) = 8
2
0
4 t 1 58. A = |x|3 e−x dx = 2 x3 e−x dx = − lim e−x 2 t→∞ 0 −∞ 0 4 1 1 1 = − lim (e−t − e0 ) = − (0 − 1) = 2 t→∞ 2 2 ∞
1
1
-2
-1
1
2
√
s
s→1
2
5 -2
7.7. IMPROPER INTEGRALS
489
1 1 √ 1 1 √ √ 60. (a) A = dx = lim + dx = lim + 2 x + 2 s→−2 s→−2 s x+2 x+2 −2 s √ √ √ = lim + (2 3 − 2 s + 2) = 2 3
4
1
2
s→−2
1
(b) V = π −2
1 dx = π lim x+2 s→−2+
1 s
1 1 dx = π lim ln |x + 2| x+2 s s→−2+
-2
2
= π lim [ln 3 − ln(s + 2)] s→−2+
The integral diverges, so the volume is infinite. 1 1 1 1 x 1 2 − dx = lim ln(x + 1) 61. A = dx = lim x x(x2 + 1) s→0+ s x2 + 1 s→0+ 2 0 s 1 1 1 ln 2 − ln(s2 + 1) = ln 2 = lim+ 2 2 2 s→0
1
2
1
∞
62. V = π
2 −2x
x e 0
t
dx = π lim
t→∞
1
2 −2x
x e
dx
0
1 u = x2 , du = 2x dx; dv = e−2x dx, v = − e−2x 2 t t 1 2 −2x −2x = π lim − x e + xe dx t→∞ 2 0 0
5 -1
1 u = x, du = dx; dv = e−2x dx, v = − e−2x 2 t t 1 −2x 1 2 −2t 1 −2x e = π lim − t e − xe + dx t→∞ 2 2 0 2 0 t 1 −2x 1 2 −2t 1 −2t = π lim − t e − te − e t→∞ 2 2 4 0 2 t2 + t π π t +t 1 1 = π lim − 2t + − e−2t = −π lim + − lim e−2t t→∞ t→∞ 2e2t 2e 4 4 4 4 t→∞ π 2t + 1 π 2 π h h = −π lim + − 0 = −π lim + = t→∞ 4e2t t→∞ 8e2t 4 4 4
CHAPTER 7. TECHNIQUES OF INTEGRATION
490
∞
63. W = 1.7×106
t 1 16 dr ≈ 4.87 × 10 lim r−2 dr t→∞ 1.7×106 r2 1 1 16 = 4.87 × 10 lim − t→∞ 1.7 × 106 t
(6.67 × 10−11 )(7.3 × 1022 )(10, 000)
= 4.87 × 10
16
lim
t→∞
1 − r
t 1.7×106
4.87 × 1016 ≈ 2.86 × 1010 joules = 1.7 × 106 64. We use the formula W = −
qq0 (a) W = − 4πe0 (b) W = −
rB rA
qq0 4πe0
rB ∞
qq0 = lim 4πe0 t→∞
qq0 4πe0
rB rA
rB 1 qq0 1 1 = − − r rA 4πe0 rB rA t t 1 qq0 qq0 1 dr = lim dr = lim − 4πe0 t→∞ rB r2 4πe0 t→∞ r rB 1 qq0 − = t 4πe0 rB
1 qq0 dr = − 2 r 4πe0 1 r2 1 rB
1 dr. r2
k k 1 e−st dt = lim e−st dt = lim − e−st k→∞ 0 k→∞ s 0 0 1 1 −sk 1 − e = lim = , s>0 k→∞ s s s
∞
∞
65. L{1} =
k 1 te−st dt = lim te−st dt u = t, du = dt; dv = e−st dt, v = − e−st k→∞ s 0 0 k k k 1 −st 1 1 1 e = lim − te−st + dt = lim − ke−sk − 2 e−st k→∞ k→∞ s s s s 0 0 0 k k 1 −sk 1 1 1 1 = lim − + 2 − 2e = − lim sk + 2 , s > 0 k→∞ s esk s s s k→∞ e s 1 1 1 1 h = − lim + 2 = 2, s > 0 s k→∞ sesk s s
66. L{x} =
k k 1 (1−s)t e et e−st dt = lim e(1−s)t dt = lim k→∞ 0 k→∞ 1 − s 0 0 1 (1−s)k 1 1 e , s>1 = lim − = k→∞ 1 − s 1−s 1−s
67. L{ex } =
68. L{e
−5x
∞
}=
∞ 0
= lim
−5t −st
e
k→∞
e
k
dt = lim
k→∞
e 0
−(s+5)t
dt = lim
k→∞
1 1 −(s+5)k 1 − e , s > −5 = s+5 s+5 s+5
k 1 −(s+5)t
e −
s+5 0
7.7. IMPROPER INTEGRALS 69. L{sin x} =
∞
491
e−st sin t dt = lim
k→∞
0
k
e−st sin t dt
0
u = e−st , du = −se−st dt; dv = sin t dt, v = − cos t k k −st −st = lim −e cos t − se cos t dt = 1 − lim s 0
k→∞
k→∞
0
k
e−st cos t dt
0
u = e−st , du = −se−st dt; dv = cos t dt, v = sin t k k −st −st = 1 − lim s e sin t + s e sin t dt = 1 − s2 L{sin x} 0
k→∞
0
Solving for L{sin x}, we have L{sin x} = 70. L{cos 2x} =
∞
e
−st
k
cos 2t dt = lim
k→∞
0
s2
1 , where s > 0. +1
e−st cos 2t dt
0
1 u = e−st , du = −se−st ; dv = cos 2t dt, v = sin 2t 2 k k s k −st 1 1 −st −st e = lim sin 2t + se sin 2t dt = 0 + lim e sin 2t dt k→∞ k→∞ 2 0 2 2 0 0 1 u = e−st , du = −se−st dt; dv = sin 2t dt, v = − cos 2t 2 k k 1 s 1 s s2 − e−st cos 2t − = lim se−st cos 2t dt = − L{cos 2x} k→∞ 2 2 2 0 4 4 0 Solving for L{cos 2x}, we have L{cos 2x} = 71. L{f (x)} =
∞
e
∞
dt = lim
k→∞
1
72. L{f (x)} =
−st
e−st e−t dt =
3
1 − e−st s
∞
s/4 s = 2 , where s > 0. 1 + s2 /4 s +4
k = 1
1 −s e , s>0 s
e−(s+1)t dt = lim
k→∞
3
−
k 1 −(s+1)t
e
s+1 3
1 −3(s+1) e = , s>3 s+1
∞
f (x) dx =
73. −∞
0
f (x) dx + −∞
= lim −e−kx t→∞
t 0
∞
t
f (x) dx = 0 + lim 0
t→∞
= lim (e0 − e−kx ) = 1 t→∞
0
ke−kx dx
CHAPTER 7. TECHNIQUES OF INTEGRATION
492
∞
t(α+1)−1 e−t dt = lim
74. (a) Γ(α + 1) =
k→∞
0
k
tα e−t dt
0
u = t , du = αt dt; dv = e−t dt, v = −e−t k k α −t α−1 −t = lim −t e + αt e dt α
α−1
0
k→∞
0
= lim (−k α e−k ) + α lim k→∞
k→∞
k
tα−1 e−t dt = lim
−
k→∞ α
0
Using repeated applications of L’Hˆopital’s Rule on kα = 0. Thus, Γ(α + 1) = αΓ(α). k→∞ ek (b) Note that
kα ek
+ αΓ(α)
k until α ≤ 0, we find that ek
lim −
∞
e
Γ(1) =
−t
k
dt = lim
k→∞
0
e−t dt = lim (−e−t ) k→∞
0
k 0
= lim (1 − e−k ) = 1. k→∞
Then, using repeated applications of part (a), Γ(n + 1) = nΓ(n) = nΓ(n − 1 + 1) = n(n − 1)Γ(n − 1) = n(n − 1)Γ(n − 2 + 1) = n(n − 1)(n − 2)Γ(n − 2) = · · · = n(n − 1)(n − 2) · · · 2 · 1 · Γ(1) = n!. −k+1 t t x−k+1 1 − x−k dx = lim = lim , k = 1 t→∞ −k + 1 t→∞ −k + 1 −k + 1 1 1 1 1 1 1 , k = 1 + = lim t→∞ 1 − k tk−1 k−1 The integral converges for k > 1 and diverges for k < 1. If k = 1 then t ∞ t 1 1 dx = lim dx = lim ln x = lim ln t, t→∞ 1 x t→∞ t→∞ x 1 1
75.
∞
1 dx = lim t→∞ xk
t
and the integral diverges. 76. Since f (x) = x2k has an infinite discontinuity at 0 when k < 0, we consider only k ≥ 0. 1 1 1 1 dx = lim dx = lim x = lim (1 − s) k=0: −∞
k>0:
s→−∞
s→−∞
s
This integral diverges. 1 x2k dx = lim −∞
s→−∞
1 s
s
x2k dx = lim
s→−∞
s→−∞
x2k+1 2k + 1
1 = lim s
s→−∞
1 x2k+1 − 2k + 1 2k + 1
This integral converges only when 2k + 1 < 0 or k < −1/2. There are no non-negative values of k for which the integral converges. t kt t ∞ e 1 kx 1 kx kx − e dx = lim e dx = lim e = lim , k = 0 77. t→∞ 0 t→∞ k t→∞ k k 0 0
7.7. IMPROPER INTEGRALS
493
∞
1 dx diverges.
The integral converges for k < 0 and diverges for k > 0. If k = 0, 0
(ln x)k has an infinite discontinuity at 1 when k < 0, we consider only k ≥ 0. x t ∞ t 1 1 k=0: dx = lim dx = lim ln x = lim ln t t→∞ 1 x t→∞ t→∞ x 1 1 This integral diverges. t ∞ t (ln x)k 1 1 k1 k+1 (ln x) dx = lim dx = lim (ln x) (ln t)k+1 k>0: = lim t→∞ t→∞ t→∞ x x k+1 k+1 1 1 1
78. Since
79.
80.
81.
82.
This integral diverges. There are no non-negative values of k for which the integral converges. ∞ ∞ 1 sin2 x sin2 x 1 dx converges. Since 0 ≤ ≤ for all x in [1, ∞), dx By Problem 75, 2 2 2 x x x x2 1 1 converges. ∞ 1 1 1 < 3 for all x in [2, ∞), dx converges. Since 0 < 3 By Example 1 in the text, 3 x x +4 x 2 ∞ 1 dx converges. x3 + 4 2 ∞ ∞ 1 1 1 1 By Problem 77, dx converges. Since 0 < < x for all x in [0, ∞), dx x x e x + e e x + ex 0 0 converges. 1, 0 ≤ x < 1 2 Let g(x) = . Then 0 < e−x ≤ g(x) for all x in [0, ∞), and −x e , 1≤x b2 2 2 2 −∞ t − 2bt + c −∞ (t − b) + (c − b ) 0 p a t−b −1 t − b =√ + lim tan √ lim tan √ c − b2 k→−∞ c − b2 k p→∞ c − b2 0 π π a aπ − − . =√ =√ 2 2 c−b 2 c − b2 √ aπ aπ c − b2 a = 8900 or = 8900. Using = 890 we obtain c − b2 c − b2 c − b2 √ 100 100 π c − b2 = 10 or c = 2 + b2 = 2 + 172 ≈ 299.13. Then a = 890(c − b2 ) ≈ 9017.59. π π (e) We note that 34 = 2b. Then the number of deaths in the first 34 weeks is We then set √
2b 0
2b
a a t−b dt = √ tan−1 √ 2 + (c − b2 ) 2 (t − b) c−b c − b2 0 b −b a =√ − tan−1 √ tan−1 √ 2 2 c−b c−b c − b2 2a b =√ tan−1 √ 2 c−b c − b2
a dt = t2 − 2bt + c
2b 0
CHAPTER 7. TECHNIQUES OF INTEGRATION
496
Using part (d) we see that the fraction of total deaths occurring in the first 34 weeks is
2b
2a b a √ tan−1 √ dt 2 − 2bt + c 2 b 2 t c−b c − b2 0∞ = = tan−1 √ . aπ a π c − b2 √ dt 2 c − b2 −∞ t − 2bt + c With b = 17 and c = 299.13 we find the percentage of total deaths within the first 34 weeks is 17 2 tan−1 √ × 100 = 88.22%. π 299.13 − 172
7.8 1.
Approximate Integration Midpoint Rule k 1 2 3 xk 3/2 5/2 7/2 f (xk ) 39/4 95/4 175/4 4 4 − 1 39 95 175 309 2 + + = 77.25 (3x + 2x) dx ≈ = 3 4 4 4 4 1 4 4 (3x2 + 2x) dx = (x3 + x2 ) = 80 − 2 = 78 1
1
2.
Midpoint Rule k 1 2 3 4 xk π/48 π/16 5π/48 7π/48 f (xk ) 0.997859 0.980785 0.94693 0.896873 π/6 π/6 − 0 (0.997859 + 0.980785 + 0.94693 + 0.896873) ≈ 0.500357 cos x dx ≈ 4 0 π/6 π/6 1 1 cos x dx = sin x = −0= 2 2 0 0
3.
Trapezoidal Rule k 0 1 2 3 4 xk 1 3/2 2 5/2 3 f (xk ) 2 35/8 9 133/8 28 3 35 133 3−1 45 ≈ 22.5 (x3 + 1) dx ≈ 2+2 + 2(9) + 2 + 28 = 8 8 8 2 1 3 4 3 x 93 5 3 +x − = 22 (x + 1) dx = = 4 4 4 1 1
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