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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