CHAPTER 4INTEGRATION

Integration is the process inverse of diff erentiation process. The integration process is used to find the area of region under the curve.

INDEFINITE INTEGRAL

Examples :

63 5 3 5 22

1. (2 ) 2 =-3 6

xx x dx x dx x dx x C

2. 3 3

= 3

x x

x

e dx e dx

e C

3

23. sin cos3

xx x dx x C

EXERCISE 4

21

32

2/3

2

1. ( 3 6)

2. 3

3. 4 2cos3 6sin 2

4. 2

5. sin

x

x

x x dx

x dx

e x x dx

dx

x dx

2

2

6. 3

1 2 37.

2

68. 2 ln 2

3 29.

10. ( 2)( 1)

x

x

e dx

dxx x x

e dxx

x xdx

x

x x dx

Examples :

THE DEFINITE INTEGRAL

22 32

1 1

1. 4 3 4 33

2 1 = 4 3(2) 4 3(1)

3 3

8 4 13 = 6 3

3 3 3

xx dx x C

C C

Integration by Substitution – Change of Variables

Reversing the “chain rule” (from diff erentiation).General integration by substitution :

Integration by Substitution steps:

Figure out the “inner” function; cal l i t u(x) . Compute . . Replace al l expressions involving the variable x and dx with the new variable u and du . Use the diff erential formula to replace the diff erential dx .

Evaluate the result ing “u” integral. I f you can’t evaluate the integral, try a diff erent choice of u .

Replace al l occurrences of the variable u in the antiderivative with the appropriate function of x .

TECHNIQUES OF INTEGRATION

'( ( )). '( ) ( ( ))f g x g x dx f g x C

du

dx

Example 1 : Calculate 3 4x dx

1 2 3 2 3 2

3 4,

3

1

31 1 1 2 2

(3 4)3 3 3 3 9

Let u x

du

dx

dx du

u du u du u C x C

Example 2:Find

43 2xx e dx

4

4

4

3

3

3 2 33

2

Let 2

4

4

41

41 1

= 4 4

x u

u

u x

u x

dux

dxdu

dxx

dux e dx x e

x

e du

e C e C

EXERCISE 5

2 3

5 2 5 2

65

1. ans: 1 ln 11

ln ln2. ans:

31

3. ans: 5

14. ans: ln 1

1

2 65. 2 6 ans:

12

6. ans: ln

x x

xx

x xx x

x x

xdx x x C

x

x xdx C

x

e dx e C

dx e Ce

xx dx C

e edx e e C

e e

43

3

2

5 3

2 2

2

sin( )7. sin( ) cos ans:

41

8. 2 7 ans: 2 731 1

9. 1 2 ans: 1 2 1 210 6

sin(tan )10. ans: cos(tan )

cos

xx xdx C

x dx x C

x xdx x x C

d C

Integration by Parts

Example 3 :Find

udv uv vdu 2 ln x x dx

2

32

3 3

3 2

3 3

3 3

ln

1

3

ln3 3

1 1 = ln

3 3

1 = ln

3 3 3

= ln3 9

u x dv x

du xv x dx

dx xdx

dux

x x dxuv vdu x

x

x x x dx

x xx C

x xx C

EXERCISE 6

2 22

3 5

2 2

22

1. ans: 2 4

2 42. 5 ans: 5 5

3 15

3. sin ans: cos sin

cos10 cos104. sin(10 ) ans: - sin10

10 50 500

5. sin

x xx xe e

xe dx C

xx x dx x x C

x xdx x x x C

x x x xx x x C

x

32 3

2 2

1 ans: cos sin

2

ln 16. ln ans:

3 9

7. ans: 2 2

x x

x x

e dx e x x C

x xx xdx x C

x e dx e x x C

Partial FractionsA quotient of polynomials : can be expressed as a simpler fraction called partial fractions.

This technique is used for rewriting problems so that they can integrate.

For example, the integral can be rewritten as using the method partial fractions. Then it is easily integrated as

( )( )

( )

P xf x

Q x

2

7

6

xdx

x x

2 1( )

3 2dx

x x

2ln 3 ln 2x x C

Types of Partial Fraction

1. Improper Fraction

2. Proper Fraction

Example 4: (Case 1)5

( 2)( 3) 2 3

( 3) ( 2) =

( 2)( 3) ( 2)( 3)

( 3) ( 2) 5

3 2 5

( ) 3 2 5

0

3 2 5

1

1

5 1 1ln 2 ln 3

( 2)( 3) 2 3

A Bdx

x x x x

A x B x

x x x x

A x B x

Ax A Bx B

x A B A B

A B

A B

A

B

dx dx x x Cx x x x

Example 5 :(Improper Fraction)3 2

2 2

22

2

2

2

5

6 61 5

= 2 61 5

= 2 ( 2)( 3)

1 =

2 ( 2) ( 3)

1 =

2

x x x xdx x dx

x x x xx

x dxx x

xx dx

x x

A Bx dx

x x

x

2

2

( 3) ( 2)

( 2) ( 3)

1 2 3 =

2 ( 2) ( 3)

1 = 2ln 2 3ln 3

2

A x B xdx

x x

x dxx x

x x x C

EXERCISE 7

6

2

2 2

2

3

7 11. ans: 5ln 3 2ln 1

( 3)( 1)

1 3 12. ans: 2 6

( 2) 12

3 53. ans: 2 3ln 1

1 2

3 34.

( 1)

xdx x x C

x x

xdx x C

x x

x x xdx x x C

x

xdx

x

22

3 2

2

2

3 2

6 ans: 3ln 1

1

5 10 15. ans: 2ln 2 5

2 5 21

6. ans: ln 2 ln 24 4

17. ans: ln

x Cx

x xdx x x x C

x x xdx

x x Cx

x xdx x

x x x

1

2C