Mathematics

Session

Definite Integrals –1

Session Objectives

Fundamental Theorem of Integral Calculus

Evaluation of Definite Integrals by Substitution

Class Exercise

Fundamental Theorem of Integral Calculus

Let F(x) be any primitive (or antiderivative) of a continuous function f(x) defined on an interval [a, b]. Then the definite integral of f(x) over the interval [a, b] is given by

b

b

aa

f x dx = F x = F b - F a

‘a’ is called the lower limit and ‘b’ the upper limit.

Note: The value of a definite integral is unique.

Iff x dx = F x +C, then

b

b

aa

f x dx = F x +C = F b +C - F a +C = F b - F a

Example - 11

0

dxEvaluate:

2x - 31

0

dxSolution: Let I =

2x - 3

1e 0

1= log 2x - 3

2

e e1

= log 2- 3 - log 0- 32

e e1

= log -1 - log -32

e e e e1 1 1

= log 1- log 3 =0- log 3=- log 32 2 2

Example - 2

4

0

Evaluate: sin3xsin2xdx

4

0

Solution: Let I = sin3xsin2xdx

4

0

1= 2sin3xsin2xdx

2

4

0

1= cosx - cos5x dx

2

Solution Cont.

4

0

1 sin5x= sinx -

2 5

5sin1 sin04= sin - - sin0-

2 4 5 5

1 1 1= +

2 2 5 2

3 3 2= =

105 2

1 1 1 1= - -

2 52 2

Example - 3

24

0

Evaluate: sin x dx

24

0

Solution: Let I = sin x dx

2 22 22

0 0

1 1= 2sin x dx = 1- cos2x dx

4 4

2 2

2

0 0

1 1 1+cos4x= 1- 2cos2x+cos 2x dx = 1- 2cos2x+ dx

4 4 2

Solution Cont.

2

0

1 1 4 sin4x= 3- 4cos2x+cos4x dx = 3x - sin2x+ 2

8 8 2 40

1 3 1 1 3 3= - 2sin + sin2 - 0- 0+0 = - 0+0 =

8 2 4 8 2 16

Example - 412

214

dxEvaluate:

x - x

1 12 2

2 21 14 4

dx dxSolution: Let I = =

1 1x - x - x - x+4 4

11

22

-1

2 21

14

4

1x -dx 2= = sin

11 1

- x - 22 2

1

-1 -1 -1214

1= sin 2x - 1 =sin 0- sin -

2

-1 1=0+sin =

2 6

Example - 5

1

0

dxEvaluate:

x+1 x+2

1

0

dxSolution: Let I =

x+1 x+2

1 A B

Let = +x+1 x+2 x+1 x+2

x+2 A+ x+1 B1

=x+1 x+2 x+1 x+2

1= x+2 A+ x+1 B Identity

Solution Cont.

Putting x =-1, - 2, we get

A =1, B =-1

1 1

0 0

dx dxI = -

x+1 x+2

1 1

e e0 0= log x+1 - log x+2

e e e e= log 1+1 - log 0+1 - log 1+2 - log 0+2

e e e=log 2 - 0- log 3 +log 2

e e e4

=2log 2 - log 3 =log3

Example - 6p

2

0

Evaluate: 3x dx =8. Find the value of p.p

2

0

Solution: We have 3x dx =8

p3

0

x3× =8

3

3p - 0=8 p=2

Evaluation of Definite Integrals by Substitution

b

a

Let I = f g x .g' x dx

Substituting g x = t g' x dx = dt

When x = a t = g a and when x = b t = g b

g b

g a

I = f t dt

Now find the result using the fundamental theorem.

Example - 7

2

3

0

Evaluate : 1+ sinx cosx dx

2

3

0

Solution : Let I = 1+ sinx cosx dx

Substituting 1+ sinx = t cosxdx = dt

When x = 0 t =1 andwhen x = t = 22

22 43

11

tI = t dt =

4

4 42 1 16 - 1 15= - = =

4 4 4 4

Example - 8a

-a

a- xEvaluate: dx

a+xa

-a

a- xSolution: Let I = dx

a+x

Putting x =acos2 dx =-2asin2 d

When x =-a = and when x =a =02

0

2

a- acos2I = × -2a sin2 d

a+acos2

0 2

2

2

2sin= -4a sin cos d

2cos

Solution Cont.0

2

sin=-4a .sin cos d

cos

02

2

=-4a sin d

0

2

01- cos2 sin2

=-4a d =-4a -2 2 4

2

= -4a 0- 0- +sin =-4a - +0 =a4 4

Example - 9

23

0

Evaluate: cos sin d

23

0

Solution: Let I = cos sin d

2 2

2 2

0 0

= cos sin .sin d = cos 1- cos .sin d

Puttingcos = t -sin d =dt

When =0 t =cos0=1 andwhen = t =cos =02 2

Solution Cont.

0

2

1

I =- t 1- t dt

01 5 3 702 2 2 2

1 1

2 2=- t - t dt =- t - t

3 7

2 2 8=- 0- 0 + - =

3 7 21

Example - 10

1

-12

0

2xEvaluate: sin dx CBSE1992, 2002

1+x

1-1

20

2xSolution: Let I = sin dx

1+x

2Putting x = tan dx =sec d

x =0 tan =0 =0 and x =1 tan =1 =4

4

-1 2

0

I = sin sin2θ sec θdθ

42

0

=2 θ sec θdθ

4

40

0

= tan - 1. tan d2

Solution Cont.

e=2 tan - 0 - -log cos4 4

40

e e1 1

=2 +log =2 - log 24 4 22

e e=2 tan - 0 + log cos - log 14 4 4

e= - log 22

Example - 11

0

1Evaluate: dx

5+2cosx

2

0

1Solution: Let I = dx

5+2cosx

2

20

2

1= dx

x1- tan

25+2x

1+tan2

2

2

2 20

x1+tan

2= dxx x

5 1+tan +2 1- tan2 2

2

Solution Cont.

2

20

xsec

2= dxx

3tan +72

2

2

20 2

xsec1 2= dx

3 7 x+ tan

3 2

2

2x xPutting tan = t sec dx =2dt

2 2

x =0 t = tan0=0 and x = t = tan =4

12

Solution Cont.

20 2

2 dtI =

3 7+t

3

1

1

3

7

-1 -1 -1

0

2 1 t 2 1= × tan = × tan - tan 0

3 7 7 3 73 3

12 3tan

7=

21

Thank you