Jee 2014 Booklet6 Hwt Solutions Integral Calculus 2

Vidyamandir Classes

Solutions to Home Practice Test/Mathematics Integral Calculus-2 1.

x sec 2   dx 2 2 x tan    9 2

 5  4 cos x   dx

Put

x tan    t 2



 5  4 cos x   t

 0

1 x sec 2   dx  dt 2 2

and

dx



2dt 2

9



2 t tan 1   3  3 

dx 2 t  tan 1   5  4 cos x 3 3



 3

0

 2

2.



I

HWT - 1

 2

 f  x   f   x  .  g  x   g   x  dx 

 2

  f  x   f   x  .  g  x   g   x    f   x   f  x  .  g   x   g  x dx 0

 2

  f  x   f   x  .  0 dx  0



0

3.

Clearly g  x   e x 1

I



f  x   x2  e x

 1



x 2  e x . e x dx 



0



1

0

0

 x2 . e x

1 0

 2  xe x   

1

1



 2 e x dx 

0

1

e2 x 2



x 2 . e x  e 2 x dx  x 2 . e x  2 x . e x dx 

0

e2 x 2

0

1 1

1

0

0

 x 2 . e x 2 xe x

 2e x

1 0



0

1

0

e2 x 2

1

0

 e2 1  e2 3   e    2e    2e  2         e   2 2 2 2  

4.

Put log x  t and e2

I1 

 e

1 dx  dt in I1 to get : x

dx  log x

30

8.

I



30

cos x dx 

30



2

 1

et dt  t

2

 1

e x dx  I2 x

 cos x  cos   x   dx   

0

  2   60  cos xdx   cos x dx   60  sin x   0   2





VMC/Integral Calculus-2

30





2 cos x dx  2  30

0

 2 0

 sin x

 cos x dx 0

   2 

 120

18

HWT-Solutions/Mathematics

Vidyamandir Classes

Integral Calculus-2 e

1.

 tan 1 x log x I    x 1  x2 1



 tan 1

2.

I



1

e

  1



e

d tan 1 xlog x dx  tan 1 xlog x dx 1

e

 sin x  1

1 

1

  dx   

HWT - 2

2

dx

sin x



 



2 2  1  sin 1 x   sin x    dx  1   sin x 1    sin x  0    1



1

  sin x  1

2

dx

0

Put x  sin to get dx  cos  d  2



I



2

cos  d

0



Now

2





cos  d   2 sin   2 sin  d   2 sin   2   cos    cos  d   

  2 sin  2 cos   2 sin

I   2 sin  2 cos   2 sin

 4.





 2 0



2  2 8 2 4 4

2 y2  1  y2 x The graph is as follows







  2 y2  y2  1 dy  4 1  dy Required area  2  2     dy  4 2 2 1 y2  1  y   1  y  0 0 0





 4 tan 1  y 

6.

 0



 2

 e 1 . e  x . e x  1  1  0 x x 2x  e , e , e  e f  x   1 x x 1  e2 x  e ,e ,e  2 x x 2  e2 x e . e . e

1  x  0 0 x 1 1 x  2 x2

The graph of f (x) is as follows

0



2



Required area = 1dx  e 2 x dx  x 1

VMC/Integral Calculus-2

0

0 1



e2 x 2

2

0

19

HWT-Solutions/Mathematics

Vidyamandir Classes  1 2 3

9.



I

e4  1 e4  1  2 2

 

Put x 3 2  t to get

x cos 2 x3 2 dx .

0



I





2 1 cos 2  t  dt  3 3

0







1  cos 2t  dt 

0

3 x dx  dt 2

1  sin 2t  t 3  2 

 0

 3

Integral Calculus-2 1.5

1.



I

2



 x 2  dx   

1

1.5

dx 

1

 2



 2

 log tan x  dx   log  cot x  dx    log tan x  dx   I

I

n

n

0

3.

 

2  1  2 1.5  2  2  2

2

 2

2.

 2dx  

Let f  x  

x2 2



t  5t  4 2e

t

0

x4  5x2  4

 f  x 

2  ex





f (x) has extremum when 2 x x 4  5 x 2  4  0

or

2 x x2  1 x2  4  0

or

x  0 ,  1,  2







6.

2

2 x  x  1 x  1 x  2  x  2   0

or

5 2

 2

 sin 1  cos x   cos 1  sin x   dx   

 sin

1

 cos x   cos 1  sin x  dx

0

 2





 2

 sin 1  sin x   cos 1  cos x   dx   

0



2 x dx  x 2

 2 0



2 4

0

x

1

0

x

 f t  dt  x   tf t  dt Diff. w.r.t. x to get f  x   1  xf  x   f  x  

 1  f  2  2 



1 3

f  2 

and

1 1 x



1 7 2  The required quadratic equation is  x  2   x    x 2  x   0 3 3 3 



3x 2  7 x  2  0





10

8.

 2x

2





 I=0

n

0

0

7.

HWT - 3

I

x  2 x  1  x  2 x  1 dx

1

Put x  1  t 2 and dx  2t dt to get : 3

3

I   t 2  1  2t  t 2  1  2t  2t dt   t  1   t  1  2t dt  





0

1

4t 3   t  1  1  t   2t dt  3

 0

VMC/Integral Calculus-2

1

3

 2t 2 1

1 0



104 110 2 3 3

20

HWT-Solutions/Mathematics

Vidyamandir Classes 9.

Let n be degree of polynomial P(x)

 

Now,

P x3  x 4 P  x 

Let

P  x   ax 2  bx  c



3n  4  n  n  2

P  0  0  c  0 P   x   2ax  b 1



P  x  dx 

P  1  7  2a  b  7

ax3 bx 2   cx 3 2

0

. . . .(i)

1

 0

a b a b 3   c  1.5    3 2 3 2 2

or

2a  3b  9

. . . .(ii)

Solving (i) and (ii), we get : a = 3, b = 1

P  x   3x 2  x



1



So,

P  x   6 x  1

and 1



P  x  . P   x  dx 

0

1

3x 2  x

0



  6x  1 dx   18x3  9x2  x  dx 0

2

9 4 x x  3 x3  2 2

1

9 1  3  8 2 2

 0

Integral Calculus-2 4.

x

xg  x   2g  2 

 3t  2 g  t  dt

. . . .(i)

   0

[See limits]

2 2

2



HWT - 4

g  2  0

Now differentiating equation (i)

6.



g  x   xg   x   3x  2 g   x 



4g  2  6  g  2 

1 n n lim

n 1





g 2  2g  2  6  2g  2

3 2

1

 k log 1    log 1  x  dx  n k 0

 0

1

 log 1  x  dx

Let 1  x  k

0



8.

0

1

1

0

1

 log k  dk    log kdk  k log k  k

 1 0

Graph is

VMC/Integral Calculus-2

21

HWT-Solutions/Mathematics

Vidyamandir Classes 1 1 1 1  2 2

Area of shaded region 

9.

log P 

  r  3 log 1        n   r 1 n



1 n

1

 log 1  x  dx 3

0

 





  13xx

1







3

log 1  x 3 dx  x log 1  x 3 

 

log 1  x 3 dx  x log 1  x 3

1 0

3

dx 1

 3x

0



0

 2

10.

 0

 1  sin 3 x   1  2 sin x  dx   

 2

 1 x

1

3dx 3

 1  2 sin x  sin x  4 sin3 x      1  2 sin x  

 0

 log 2  3  3 0

[Using ( sin 3 x  3 sin x  4 sin3 x) ]

 2



 1  sin x 1  2 sin x   1 0

Integral Calculus-2 1.

Let

2

2

0

0

HWT - 5

 f t  dt  A,  tf t  dt  B f  x   2  B  x2 A



2

2

2

2

 f  x  dx   2   B dx   x A dx  A



2

0

0

2



and

xf  x  dx 

0

2

2

3   2  B  x dx   x A dx  B

0

. . . .(i)

0

0

. . . .(ii)

0

Solve (i) and (ii) to get : A

12 28 ,B 19 19

f  x  



12 2 10 x  19 19

1

 f  x  dx  19



6

0

4.

Since x3  x 4 for 0 < x < 1 

I1  I 2

Since x3  x 4 for 1 < x < 2 

I 4  I3  2

5.

A B 



 sin x  cos x 

  sin x  cos x  dx  2 0

 2

A

 0

 2

sin x dx  sin x  cos x

 0

  sin   x  dx 2       sin   x   cos   x  2  2 

 2



 cos x  sin x = B cos x dx

0

Since A = B



AB

VMC/Integral Calculus-2

 4

22

HWT-Solutions/Mathematics

Vidyamandir Classes 1

7.

 1  x 

 x  1  1 e x dx   e x  e x  0  x  12 0  1  x 1  x 2  1

xe x

2

dx 

0

1

 2

  sin

8.

dx 

 2

2

x  sin 4 x

 2

  sin x  cos x  dx    sin

2

x  sin 4 x

ex 1 x

1

 0

e 1 2

  2 cos x  dx

0

Let sin x  t 1

 t



2

t

4



2t 5 2t 3  2  dt   5 3

0

1

 0

2 2 2  3  5  4    5 3 15 15

Integral Calculus-2  4

1.

 0

 cos 2 x dx     1  sin x     4





 4

 0

cos 2 x dx 1  sin x

cos 2 x  2  dx 2



cos x

0

HWT - 6

 2

3.

1

 x3   2 2 1 Shaded Area  2   x dx  2 2 tan x      1  x2  3  

1

 0

0

  1  3  2  2    3  2 3 4

5.

Shaded area = 2



x 2  4 dx

2

4

 x x2  4 4   2  log  x  x 2  4      2 2  2









 2  2 2 3  2 log 4  2 3  2 log 2   









x  dx 

16 I 3

 2 4 3  2 log 2  3   8 3  4 log 2  3   4

7.

4

  x  dx    0

0



2 x   x  dx  x 3 2 3

4

4

 0

  0

4

I

 

x  dx

Critical points are where

x I

0 1



x  I2





4



I  0 dx  1dx  3 0

1

4



16 7   x  dx  3  3  3 0

VMC/Integral Calculus-2

23

HWT-Solutions/Mathematics

Vidyamandir Classes 9.

f (x) is

 2

 1  cos x  dx 

Find area of shaded region which is area 

0

bc

10.



f  x  c  dx

5 6



1 dx 

 2





2 sin x dx

5 6

xc t

ac

bc



dx  dt





f  x  c  dx 

ac

b

 f t  dt a

Integral Calculus-2 12

1.



1 2

2

HWT - 7

2

 x 1   x 1   x  1    x  1   2 dx     12



2

0

2

12

2

  x  1 4x

2

12

dx  2

0

12

2

4 x

x 0

12

2

 x 1   x 1   x  1    x  1   2 dx  2    

2

1

4 x

x 0

2

1

 x 1   x 1 

  x  1    x  1  dx 0

dx

 4x  1  is  ve in x   0 ,    As 2  2  x 1 

12

dx  4

x

2x 2

0

1

dx  4n x 2  1

12 0

 3   4  n  n1  4 

 4n  4 n

2.

3 4

4 3

y 2  4ax x 2  4ay 4a

Area 

 0

  2 4a x 3 2 x2  x3     4ax   dx   4a  3 12a   

4a

0

 32a 64a     3 12   2

2

 32  16  2  a  3  

VMC/Integral Calculus-2

16a 2 3

24

HWT-Solutions/Mathematics

Vidyamandir Classes  4

2

5.

 0



 cos x  sin x  dx 

 sin x  cos x  dx 

 4

0

1

6.

5 4



cos x  sin x dx 

 x 1  x 

99

dx 

 1  x  1  1  x 

99

1

5 4

dx

1

99 99 100  1  x  x dx    x  x  dx 0



0

100

101

x x  100 101 4

1

 0

1 10100



1

 x tan  x  x  dx  1 dt  t2

1

4

1

14 14

 t tan  t  t

I  I



cos x  sin x dx

0



14



1

0

8.

2



Let

4

1



1

 t tan  t  t  dt    t tan  t  t  dt 1

4



1

1 t x

14

I=0

Integral Calculus-2 /2

4.(C)



Let I 

/2

dx 1  cot x

0

/2

Also I 

 0





0

dx /2

sin  / 2  x  dx

sin  / 2  x   cos  / 2  x  /2







0

cos x cos x  sin x

dx



 1dx  2 or I  4

2I 



sin x sin x  cos x

HWT - 8

0



5.(C)

Let I 



xdx

0

 x  dx

0







 4 cos2 x  9 sin2 x  4 cos2 x  9 sin2 x 

2I 

/2

dx

 4 cos2 x  9 sin2 x  2  0

0

/2

 2

 0



6.(B)

2 9



sec 2 xdx 2

4  9 tan x



dt

 t2  4 / 9 0



 



 2

2

dx 2

4 cos x  9 sin 2 x dt

 4  9t 2 , where tan x  t 0

3

. tan 9 2

1  3 

 t  2 







2 6

0



 



Let F  x   f  x   f   x  . g  x   g   x  then F   x   f   x   f  x  . g   x   g  x    F  x  /2





F  x  dx  0

 / 2



7.(A)

dx

  x2  4  x2  9



0



 1 1    dx  5  x 2  4 x 2  9  0

1





1 1 1     x 1  x    tan 1    tan 1         5  2 2 3 3      0 5  4 6  60

VMC/Integral Calculus-2

25

HWT-Solutions/Mathematics

Vidyamandir Classes /4

8.(A)

Required area 



 cos x  sin x  dx 

/2



 sin x  cos x  dx

/4

0

/4 /2 =  sin x  cos x     cos x  sin x  2 0 /4 x

9.(C)

2  t  1 t  2 dt

y

dy



dx

0

d2y dx

2

 0  x  2

2 / 4

10.(C)

I



sin

d2y

  x  1  x  2   2

dx 2





2 1

  x  2   3x  4 

4 3 /2

 



x dx . Put x  t 2 to get I  2

0

t sin  t  dt

0

/2  2 t cost  sint  2 0

Integral Calculus-2

1.(D)

2.(D)

3.(B)

Put 3x  t to get I1 

1

12  12T



3

2  2 1 x  1  dx  xdx  x 

f  t  dt 

12





0

0

1

3 

3



x 2dx 

f  t  dt 

1 3

12 I   4 I

2

 x dx

x3dx 

4

2





12T

0

2



1

HWT - 9

3





1 1 3/ 2 1 1  2  1   9  4   32  35 / 2 2 3 4 5



469 1 3 / 2 1 5 / 2  .2  .3 60 3 5



The curves y  5  x 2 and y  |x  1 | intersect at (2, 1) and (–1, 2). 2

Required area 



2



5  x 2 dx  |x  1 |dx

1

1 2

2

x  x  5 5  x dx   5  x 2  sin 1    2  5  2 1 1



2

2

5 4

2



1

|x  1| dx 

1

2 0



1

  t  dt  

2



0

t 2 t dt  2

0

Required area 

 f  x  2x e



2  x 2 4

4.(A)

 | t | dt , where t  x  1  2

t2 2

1

 0

5 2

5 1  . 4 2

 2 xe

 x4

 2 xe





2  x2  4 

16  8 x 2  1 e   

2

Since : 1  e16  8 x  0 for all x, f increases for x < 0

VMC/Integral Calculus-2

26

HWT-Solutions/Mathematics

Vidyamandir Classes /4

5.(C)

I

/4



 



log 1 tan 2   2 tan d  2

0

log 1  tan  d

0 /4



2



/4





log 1  tan  / 4    d  2

0

0

/4



2

 1  tan  log 1  d  1 tan 

 log  2   log 1 tan   d  log  2   I 2



I

 log  2  4

0 3

6.(C)



f  x  dx 

2

2



f  x  dx 

2

3



2

f  x  dx 



3

ecos x . sin xdx 

2

2

 2dx 2

a

 0  2  2 since

 g  x  dx  0 if g   x    g  x 

a 3 / 4

7.(A)

I



/4

3 / 4



dx  1  cos x

2I 



/4

8.(D)



  3    x  / 4 1  cos  4 4 

3 / 4



3 / 4

dx

  1 1    dx  1  cos x 1  cos x  



/4

dx 1  cos x 3 / 4

3 / 4



/4

2 1  cos 2 x

dx   2 cot x

4  I  2 /4

The points of intersection of the two curves are  21 and  2  1 . 1

Required area 



2 2  | 3 y    2 y  dy

1

y3 3

y

9.(C)

Let f   x   6  x 2

10.(B)

 1

4 3

x2  f   x   6  x2  x4  6  x2



x4  x2  6  0 

or

1

 x  3  x  2   0 2

2

x2  2  0  x   2



The function g  x   | x | is an even function while the function h  x   f  x  cos x is an odd function. 

1

1

1

1

0

0

 |x|  f  x  cos x  dx  2|x|dx  2 xdx  1 Integral Calculus-2

/4

1.(C)

I



sin x  cos x dx   3  sin 2 x

0

/4

 0

 sin x  cos x  dx  sin x  cos x 2  4 0

Put sin x  cos x  t to get I  



1



HWT - 10

1  t 2   log 2 t 2 t  4 4  dt

  

0

1

1 1  log 1 log 3  log  3 4 4

VMC/Integral Calculus-2

27

HWT-Solutions/Mathematics

Vidyamandir Classes 



Required area 

2.(D)

|cos x|dx

5 / 6  / 2





/2



cos x dx 

5 / 6

 / 2

C1  C1  C2  C3  f  x   f  x  

3.(D)



cos x dx 

cos x dx   sin x

/2

sin x 0 0

 sin x 5 / 6



/2

 / 2





 sin x  / 2

7 . 2

/2

sin 2 x sin 3x 3 4 sin x sin x 1





 sin x 3  4 sin 2 x  sin 3x /2





f  x  dx 

/2



0

0

x2

   f t  dt 

F x2 

4.(C)

/2

1 sin 3x dx  cos 3 x 3

1 3

 0

x 2 1 x  

0

x2

 f t  dt 0

 

Differentiate both sides to get 2 x  3 x 2  2 x . f x 2

 

f x 2 1 



3x  f  4  1  3  4 2

f   x   4 x3  6 x 2  2 x  2 x  2 x  1  x  1

5.(B)

f   x   0  x  0 1

Required area 





1  1 . Note that f  x  has minimum at x  01 2



x5 x 4 x3 x  2 x  x  3 dx     3x 5 2 3 4

3

2

0



0

1 1 1 91   3 5 2 3 30

2

2

6.(C)

1

 f   x  dx  f  x  0

 f  2  f  0  7  4  3 0

x

 sint dt 2

7.(B)

lim 0

x 0

sin x

2

 

x 0 cos x 2 . 2 x

0

8.(D)

sin x 2

 lim

Required area 



 0

0

0

r2

n

 r3  n3  nlim  n   n lim

r 1



1

1

3 4 3 4 x3 3  x 3  x 3   4 4 3 2 1 0 1 0 n

1

2 x2

1   2  1 2  x  x 3  dx   x 3  x  dx    

x3  3

Required limit =

x 0

  .x0

tan x 2

1

1

9.(B)

 lim

x2

1

r n  

r 11

2

r n  

3

 1  x3 dx  3 log  2 1

0

VMC/Integral Calculus-2

28

HWT-Solutions/Mathematics

Vidyamandir Classes sec 

10.(D)

I



f  x  dx  

cos 

sec 



f 1 / x 

cos  cos 

I



f  t  dt 

sec 

VMC/Integral Calculus-2

x

2

cos 



dx . Put

1  t to get x

f  x  dx   I



2I  0

 I 0

sec 

29

HWT-Solutions/Mathematics