Integration

CHAPTER 3: INTEGRATION 3.0 Basic Concepts Integration is the reverse process of differentiation 1. Given

, therefore ∫

2. Given

, therefore

1. Given

Find

.

,

2.

.

Given Find

, .

Solution :

3.

Given Find

5.

, .

Given

Find

Given

Find

4.

,

6.

.

,

.

Given

,

Find

.

7.

7.

Given Find

, .

8.

Given Find

, .

9.

Given Find

, .

10. Given

Find

, .

3.1. Indefinite Integral

3.1.2 (A) Integration of

Exercise : 1. Determine the following integrals ;

1.

2.

3.

4.

6.

7.

8.

=

=

5.

=

=

3.1.2 (B) Integration of

'a' is a constant

Exercise : 2. Determine each of the following integrals ;

1.

2.

3.

6.

7.

8.

10.

11.

12.

4.

=

5.

= = = =

9.

= = = =

3.1.3. Integration of Algebraic Expression

Application : Basic techniques of integration on each term. Expand and simplify the given expression where necessary.

e.g.

=

1.

5.

2.

6.

3.

4.

7.

8.

9.

10.

13.

14.

11.

12.

15.

16.

3.1.4 (A) Determine the Constant of Integration

e.g. 1. Given that

2. Given that

Find y in terms of x.

Find y in terms of x.

Solution ;

hence

3. Given that

Find y in terms of x.

4. Given that

5. Given that

6. Given that

Find y in terms of x.

Find y in terms of x.

Find y in terms of x.

7. Given that

8. Given that

9. Given that

Find p in terms of v.

Find y in terms of x.

3.1.4 (B) Determine the Constant of Integration

Find in terms of x if y=12 when x=5

e.g. 1. Given

2. Given

3. Given

Find the value of y when x=3

Find the value of y when x= -1

Find the value of y when x= 0

Solution ;

-3

3

4. Given

5. Given

6. Given

Find the value of y when x= 2

Find the value of y when x= 3

Find the value of y when x= -1

-7

25

8/3

3.1.5. Determine the equation of curve from gradient function

e.g. 1. The gradient function of a curve which

2. The gradient function of a curve which

passes through A(1,5) is

passes through B(1,-3) is

equation of the curve. Solution :

. Find the

equation of the curve.

. Find the

3. The gradient function of a curve which

4. The gradient function of a curve which

passes through P(2,3) is

passes through Q(1,6) is

. Find the

. Find the

equation of the curve.

equation of the curve.

5. The gradient function of a curve which

6. The gradient function of a curve which

passes through R(-1,4) is

passes through S(1,10) is

equation of the curve.

. Find the

equation of the curve.

. Find the

7. The gradient function of a curve which passes through X(-1,7) is the equation of the curve.

8. The gradient function of a curve is . Find

. Find the equation of the curve if it passes through the point (-2,3)

3.1.6. Integration of expressions of the form Using formula :

+ c , n ≠ -1

e.g. 1.

2.

3.

4.

5.

6.

+c = 7. 9.

3.2 Definite Integral

2.

e.g. 1.

3.

52 4.

6.

5.

20

64

3/2

8/9

8.

7.

9

-10

22.5

25/3

3.2.1 Definite Integral of the form

e.g. 1.

2.

3.

203/4

4.

5.

1280 7.

24

6.

5/12 8.

-65/3 9.

1

5

3.75

3.2.2(A) Application of Definite Integral

Exercise : 1. Given

a.

b.

c.

12

8

-6

d.

14 Exercise : 2. Given Find the value of ;

f.

e.

7

44

a.

b.

c.

-4

8

20 f.

d.

e.

0

6 16

Exercise : 3. Given Find the value of ;

b. 3

a.

10

30

c.

2

d.

5

15/2

3.2.2 (B) Applications of Definite Integral

1.

Given that Where k >-1 , find the value of k . Solution ;

2.

Given that Where k >0 , find the value of k .

4

3.

Given that

4.

Where k >0 , find the value of k .

Given that

Where k >0 , find the value of k .

4

2

5.

Given that

6.

Given that find the value of k

find the value of k . Solution ;

; k = 1/4 2 7.

Given that

8.

Given that find the value of k

find the value of k .

8/21 -2

3.3 Area under a curve using Definite Integral

3.3.1 Area under a curve bounded by x-axis y Note : 1. Area above the x-axis has a positive value. 2. Area beneath/below the x-axis has a negative value. 0

a

b

x

Exercise : 1. Find the area of the shaded region.

a)

b)

y

y

-3 0

1

x

2

x

3

Solution:

= 20

c)

d)

y

0

1

5

x

y

0

5

x

d)

e)

y

y 0

2

0

x

4

x

3.3.2 Area under a curve bounded by y-axis y

b

Note : 1. Area on the right side of the y-axis has a positive value.

a

2. Area on the left side of the y-axis has a negative value.

0

x

a)

y

y 2

2 x 2

x

0

Solution:

10

b)

d)

y

2

2

0 1

y

0

x

2

6

x

e)

f)

y

y 1 9

0 1

4 0

x

x

3.3.3. Area between two graphs y

0

a

x

b

Exercise : 3. Find the area of the shaded region a)

b) y

y

x

3

0

0

y 1

0

1

x

x

c)

d)

y

0

y y=f(x)

3.4 Volume of Revolution A.

0

a

b

x

Region bounded by a curve when is rotated completely about x-axis.

x

Volume of shaded region =

Exercise 1 : Find the volume of solid generated when the shaded region revolves 360 o about x-axis. a)

b)

y

x

2

0

c)

y

1

0

y

3

0

x

6

y

3

x

0

1

3

x

B. Region bounded by a curve when is rotated completely about y-axis y b Volume of shaded region =

a x

0

Exercise 2 : Find the volume of solid generated when the shaded region revolves 360 o about y-axis. a)

b)

y

y

3

1

0

x

0

x

y

y

c)

d) 1 1 0

0 x

-

x

Extra exercises : 1)

y 6 4

2

-2

3

x

The diagram shows part of the curve y = 4–x2 and a straight line y + 2x = 6. Find the volume generated when the shaded region is rotated through 360o about x-axis. Solution : y + 2x = 6 y = 6 – 2x Volume of a cone

2)

y

2

1

x

Find the volume generated when the shaded region is rotated through 360o about the xaxis.

3) y

=36

x

1

Find the volume of solid generated when the shaded region is revolve 360o about x-axis.

4)

y

0

x

Find the volume of solid generated when the shaded region is revolved 360o about y-axis.

5)

y 2 x -2

Find the volume of solid generated when the shaded region is revolved 360o about y-axis.

6)

y

0

x

Find the volume of solid generated when the shaded region is revolved 360o about y-axis.