mathematics. session definite integrals –1 session objectives fundamental theorem of integral...
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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