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1. DEFINITE INTEGRALS
DEFINITE
INTEGRALS

2. DEFINITE INTEGRALS

3. DEFINITE INTEGRALS
Introduction
 The notation for definite integral 𝒂
𝒃
𝒇 𝒙 𝒅𝒙 was proposed
by Jean Baptiste Joseph Fourier (1768-1830) and
Cauchy immediately adopted and popularized it.
Augustin Louis Cauchy
(1789-1857)

4. DEFINITE INTEGRALS
 In this chapter we discuss the definite integral,
interpretation of definite integral as an area,
fundamental theorem of integral calculus, properties of
integrals, reduction formulae.

5. DEFINITE INTEGRALS
Definite Integral
 A function f:[a,b]R is integral on [a,b] and
𝒇 𝒙 𝒅𝒙 = 𝑭(𝒙)+c
 The real number ‘a’ is called lower limit and ‘b’ is
called upper limit of the definite integral.
Then F(b)-F(a) is called definite integral of f(x) over [a,b]
and it is denoted by 𝒂
𝒃
𝒇 𝒙 𝒅𝒙

6. DEFINITE INTEGRALS
INTERPRETATION OF DEFINITE INTEGRAL AS AN AREA
 𝒂
𝒃
f x dx represents the area of the region bounded by
the curve y=f(x), x-axis and ordinates x=a, x=b
y
O
x
y=f(x)
x=a x=b

7. DEFINITE INTEGRALS
FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS
 If f:[a,b]R and ‘F’ is a primitive of ‘f’ then
𝒂
𝒃
f x dx = 𝑭 𝒃 − 𝑭(𝒂)
Note f(x)dx = F(x)+c then
a
b
f(x) dx= F(x)+c 𝒂
𝒃
= (F(b)+c)-(F(a)+c)
= F(b) - F(a)

8. DEFINITE INTEGRALS
Example (i)
0
2
x2dx =
x3
3 𝟎
𝟐
=
1
3
x3
𝟎
𝟐
=
1
3
[23 - 03]
=
8
3

9. DEFINITE INTEGRALS
Example (ii)
0
1
1
1+x2
dx
= Tan−1x 𝟎
𝟏
= Tan-1(1)-Tan-1(0)
=

4
- 0
=

4

10. DEFINITE INTEGRALS
THEOREM 1:
PROOF:
a
𝒃
f(x) dx =
a
𝒃
f(t) dt
Let  f(x) dx = F(x) then  f(t) dt = F(t)
a
𝒃
f(x) dx = F(x)+c 𝒂
𝒃 = F(b) – F(a)
and
a
𝒃
f(t) dt = F(t)+c 𝒂
𝒃
= F(b) – F(a)

a
𝒃
f(x) dx =
a
𝒃
f(t) dt

11. DEFINITE INTEGRALS
THEOREM 2:
If f(x) is an integral function on [a,b] and g(x) is derivable
on [a,b] then
a
𝒃
(fog)(x)g1(x) dx =
g(a)
𝒈(𝒃)
f(x)dx
PROOF:
L.H.S

a
𝒃
(fog)(x)g1(x) dx =
a
𝒃
f(g(x)) g1(x) dx

12. DEFINITE INTEGRALS
Sub g(x)=t then g′(x)dx=dt
Also if x=a then t=g(a)
and x=b then t=g(b)
=
g(a)
𝒈(𝒃)
f(t) dt =
g(a)
𝒈(𝒃)
f(x)dx
∵
a
b
f(x) dx =
a
b
f(t) dt

a
𝒃
(fog)(x) g1(x) dx =
g(a)
𝒈(𝒃)
f(x)dx

13. DEFINITE INTEGRALS

14. DEFINITE INTEGRALS
1 i)
1
2
x5dx
Solution:
1
2
x5dx=
x6
6 𝟏
𝟐
∵ xndx =
xn+1
n+1
+c
=
1
6
x6
𝟏
𝟐 =
1
6
26−16
=
1
6
[64-1] =
63
6 =
21
2
Evaluate the following integrals

15. DEFINITE INTEGRALS
ii)
0
a
(a2x−x3)dx
Solution:
0
a
(a2x−x3)dx =
0
a
a2xdx −
0
a
x3dx
= a2 x2
2 𝟎
𝒂
-
x4
4 𝟎
𝒂
= a2 a2
2
−0 -
a4
4
−0
=
a4
2
−
a4
4
=
a4
4

16. DEFINITE INTEGRALS
2 i)
0
4
x2
1+x
dx
Solution:
0
4
x2
1+x
dx
=
x2
2
−x+log(1+x
𝟎
𝟒
0
4
x2−1+1
1+x
dx =
0
4
x−1+
1
1+x
dx
=
42
2
−4+log(1+4) −(0−0+log1)

17. DEFINITE INTEGRALS
= 8-4+log5
= 4+log5

18. DEFINITE INTEGRALS
2 ii)
0
a
a− x
𝟐
dx
Solution:
0
a
a− x
𝟐
=
0
a
a+x−2 a x dx
= ax +
x2
2
−2 a
x3/2
3
2 𝟎
𝒂
= a(a)+
a2
2
−2 a
a3/2
3
2
− [0+0-0]

19. DEFINITE INTEGRALS
= a2+
a2
2
−
4a2
3
=
6a2+3a2−8a2
6
=
a2
6

20. DEFINITE INTEGRALS
3 ii)
0
π/4
sec4x dx
Solution:
=
0
π/4
sec2x.(1+tan2x)dx
=
0
π/4
(sec2x+sec2xtan2x)dx
f(x) n f1(x) dx=
f(x)
n+1
n+1
+ c
0
π/4
sec4x dx =
0
π/4
sec2x.sec2x dx

21. DEFINITE INTEGRALS
= tanx 𝟎
π/4
+
tan3x
3 𝟎
π/4
=(1−0) +
1
3
−𝟎
= 1+
1
3
=
4
3

22. DEFINITE INTEGRALS
3 iii)
0
π
2+2cosd
Solution:
0
π
2+2cosd
= 0
π
2.2cos2 
2
𝒅
=
0
π
2.cos

2
𝒅
= 2
sin
2
1
2 𝟎

=
0
π
2(1+cos)d

23. DEFINITE INTEGRALS
= 4 sin

2 𝟎
π
= 4(1-0)
= 4
= 4 sin
π
2
− sin
0
2

24. DEFINITE INTEGRALS
4 i)
2
3
2x
1+x2
dx
Solution:
2
3
2x
1+x2
dx = log(1+x2) 𝟐
𝟑
= log(1+32)-log(1+22)
= log10-log5
= log
10
5
= log2
∵
f1(x)
f(x)
dx=log f(x) +c

25. DEFINITE INTEGRALS
4 ii)
0
a
1
x2+a2
dx
Solution:
0
a
1
x2+a2
dx =
1
a
tan−1 x
a 𝟎
𝒂
=
1
a

4
−𝟎
=

4a
=
1
a
tan−1 a
a
−tan
0
a

26. DEFINITE INTEGRALS
4 iii)
0
1
x2
1+x2
dx
Solution:
0
1
x2
1+x2
dx =
0
1
x2+1−1
1+x2
dx
=
0
1
x2+1
1+x2
−
1
1+x2
dx
=
0
1
1−
1
1+x2
dx
= x−tan−1x 𝟎
𝟏

27. DEFINITE INTEGRALS
= [1-tan-1(1)] – [0-tan-1(0)]
= [1-

4
]-0
= 1-

4

28. DEFINITE INTEGRALS
4 iv)
0
3
x
x2+16
dx
Solution:
0
3
x
x2+16
dx =
1
2 0
3
2x
x2+16
dx
=
1
2
2 x2+16 𝟎
𝟑
= x2+16 𝟎
𝟑
= 32+16 - 0+16
= 25 - 16 = 5-4 = 1
∵
f1(x)
f(x)
dx=2 f(x)+c

29. DEFINITE INTEGRALS
4 vi)
0
2
4−x2dx
Solution:
0
2
4−x2dx =
0
2
22−x2dx
=
x
2
22−x2 +
22
2
sin−1 x
2 𝟎
𝟐
= 0+
4
2
sin−1 2
2
-[0+0]
= 2sin-1(1) = 2

2
= 
∵ 0
a
a2−x2dx
=
x
2
a2−x2+
a2
2
sin−1 x
a
+c

30. DEFINITE INTEGRALS
Sub 3-2x=t
-2dx=dt
Lower limit: If x=0 then
Upper limit: If x=1 then
5 i)
0
1
1
3−2x
dx
Solution:
0
1
1
3−2x
dx
 dx= -
1
2
dt
t=3-2(0)=3
t=3-2(1)=1

31. DEFINITE INTEGRALS
= −
1
2 3
1
1
t
dt
= −
1
2
2 t 𝟑
𝟏
= − 1 − 3
= 3 − 1
=
3
1
1
t
−1
2
dt

32. DEFINITE INTEGRALS
5 iii)
0
16
x
𝟏
𝟒
1+x1/2
dx
Solution:
0
16
x
𝟏
𝟒
1+x1/2
dx=
0
16
x
𝟏
𝟒
1+ x1/4 𝟐
dx
Sub x1/4=t Then
1
4
x-3/4dx = dt
dx = 4x
𝟑
𝟒 dt
dx = 4t3 dt
If x=0 then t=01/4 = 0
If x=16 then t=161/4 = 2

33. DEFINITE INTEGRALS
=
0
2
t
1+t2
(4t3dt)
= 4
0
2
t𝟒
1+t2
dt
= 4
0
2
t𝟒
−1+1
1+t2
dt
= 4
0
2
t2−1+
1
1+t2
dt
= 4
t3
3
−t+tan−1(t)
𝟎
𝟐

34. DEFINITE INTEGRALS
= 4
8
3
−2+tan−1(2)−0
= 4
2
3
+tan−12

35. DEFINITE INTEGRALS
Sub x2 = t
2x dx=dt
If x=0 then t=02=0
If x=1 then t=12=1
6)
0
1
x.e−𝒙𝟐
dx
Solution:
0
1
x.e−𝒙𝟐
dx
 x dx =
1
2
dt
=
0
1
e−𝒕
dt
2

36. DEFINITE INTEGRALS
=
1
2 0
1
e−𝒕dt
=
1
2
−e−t
𝟎
𝟏
=
−1
2
1
e
−1
= −
1
2
e−1−𝒆𝟎

1
2
1−
1
e

37. DEFINITE INTEGRALS
7)
0
/2
x sinx dx
f=x, g=sinx
Solution:
0
/2
x sinx dx
∵f.gdx=fg dx- (d(f)  g dx)dx
= x
0
/2
sinx dx −
0
/2
d x
0
/2
sinx dx dx

38. DEFINITE INTEGRALS
= −x cosx 𝟎
/2
−
0
/2
(1)(−cosx) dx
= 0−0 +
0
/2
cosx dx
= sinx 𝟎
/2
= 1
= sin
π
2
− sin𝟎
= 𝟏 − 𝟎

39. DEFINITE INTEGRALS

40. DEFINITE INTEGRALS
1
0
π/4 sinx+cosx
9+16sin2x
dx
Solution:
0
π/4 sinx+cosx
9+16sin2x
dx
Substituting sinx-cosx=t
If x=0 then t=sin0-cos0=-1
If x=

4
then t=sin

4
-cos

4
=0
Let sinx-cosx=t
Squaring on
both sides
Sin2x+cos2x-2sinxcosx=t2

41. DEFINITE INTEGRALS
1-t2 = sin2x
=
−1
0
1
9+16(1−t2)
dt
=
−1
0
1
25−16t2
dt
=
1
16 −1
0
1
25
16
−t2
dt
=
1
16 −1
0
1
5
4
𝟐
−t2
dt

42. DEFINITE INTEGRALS
=
1
16
1
2 5
4
log
5
4
+t
5
4
−t
−𝟏
𝟎
=
1
40
log
5
4
+0
5
4
−0
−log
5
4
−1
5
4
+1
=
1
40
log1−log
1
9
=
1
40
0−log
1
9
∵ 
1
a2−x2 dx=
1
2a
log
a+x
a−x
+c

43. DEFINITE INTEGRALS
=
1
40
log9
=
1
40
log32
=
1
40
2log3
=
1
20
log3

44. DEFINITE INTEGRALS
2.
0
π/2 1
4+5cosx
dx
Solution:
0
π/2 1
4+5cosx
dx
Substituting tan
x
2
=t
then dx=
2dt
1+t2
,
Cosx =
1−t2
1+t2
If x=0 then t=tan
0
2
=0
If x=

2
then t=tan

2
2
=1

45. DEFINITE INTEGRALS
=
0
1
1
4(1+t2)+5(1−t2)
(1+t2)
.
2dt
(1+t2)
= 2
0
1
1
4+4t2+5−5t2
dt
= 2
0
1
1
9−t2
dt
= 2
0
1
1
32−t2
dt
= 2
1
2 3
log
3+t
3−t 𝟎
𝟏
∵ 
1
a2−x2 dx=
1
2a
log
a+x
a−x
+c

46. DEFINITE INTEGRALS
=
1
3
log
3+1
3−1
− log
3+0
3−0
=
1
3
log2− log1
=
1
3
log2

47. DEFINITE INTEGRALS
3 (i)
a
𝑏
(x−a)(b−x) dx
Solution:
Substituting x=acos2+bsin2
dx=[-2acossin + 2bsincos] d
 dx=2sincos [b-a] d
 dx= [b-a] sin2.d
Let x-a = acos2+bsin2-a
= -a (1-cos2)+bsin2
x-a = sin2(b-a)

48. DEFINITE INTEGRALS
Let b-x=b-acos2-bsin2
=b(1-sin2)-acos2
b-x=cos2(b-a)
If x=a then cos2(b-a) = (b-a)
cos2 = 1
 =0
If x=b then cos2(b-a) = (b-b)
cos2 = 0
 =

2

49. DEFINITE INTEGRALS
Let
a
𝑏
(x−a)(b−x) dx
=
0
π/2
sin2(b−a)cos2(b−a). sin2(b−a)d
=
0
π/2
(b−a)sincossin2(b−a)d
= (b−a)𝟐
0
π/2 2sincos
2
sin2d
=
(b−a)𝟐
2 0
π/2
sin2.sin2d
=
(b−a)𝟐
2 0
π/2
sin22d

50. DEFINITE INTEGRALS
=
(b−a)𝟐
2 0
π/2 1−cos4
2
d
=
(b−a)𝟐
4 0
π/2
(1−cos4)d
=
(b−a)
𝟐
4
−
sin4
4 𝟎
/𝟐
=
(b−a)
𝟐
4

2
−0−(0−0)
= (b-a)2 
8

51. DEFINITE INTEGRALS
4.
0
1/2
xsin−1x
1−x2
dx
Solution:
0
1/2
xsin−1x
1−x2
dx
Substituting sin-1x=t  x=sint

1
1−x2
dx=dt
If x=0 then t=sin-1(0)=0
x =
1
2
then t=sin-11
2
=

6

52. DEFINITE INTEGRALS
0
1/2
xsin−1x
1−x2
dx =
0
/6
t sint dt
f(x)=t , g(x)=sint
= −t cost 𝟎
/𝟔
−
0

6
(1)(−cost)dt
∵f(x).g(x) dx=f(x)g(x) dx
- [f1(x)  g(x) dx]dx
= −

6
cos

6
− 𝟎 +
0

6
cost dt
= t
0
/6
sint dt−
0

6
d t
𝟎

6
sintdt dt

53. DEFINITE INTEGRALS
= −

6
3
2
+ sint 𝟎
/𝟔
= -
3
12
+ sin

6
−sin0
= -
3
12
+
1
2
=
1
2
-
3
12

54. DEFINITE INTEGRALS

55. DEFINITE INTEGRALS
1)
0
𝒂
kx dx
a) k b)
ka2
2
c)
ka2
3
d)
ka
2

56. DEFINITE INTEGRALS
2)
0
𝟏
1
1+x2
dx
a)

3
b)

2
c)

4
d)

6

57. DEFINITE INTEGRALS
3)
1
𝟐
1
x
dx
a) logx b) log2 c) log3 d) 0

58. DEFINITE INTEGRALS
4)
0

sinx dx
a) 1 b) -2 c) 2 d) -1

59. DEFINITE INTEGRALS
5) If f(x)dx=F(x)+c then
a
b
f(x) dx=
a) F(a) b) F(b) c) F(a)-F(b) d) F(b)-F(a)

60. DEFINITE INTEGRALS
6)
0
𝟏
ex dx
a) e b) e-1 c)
1
e d) 0

61. DEFINITE INTEGRALS
7) If
1
𝟏𝟎
f(x)dx=100 then
1
𝟏𝟎
f(t)dt =
a) 99 b) 101 c) 0 d) 100

62. DEFINITE INTEGRALS
8)
0
/𝟒
(tan4x+tan2x)dx =
a) 1 b)
1
2 c)
1
3 d)
1
4

63. DEFINITE INTEGRALS
9)
0

(a−x−b−x) dx =
a)
1
loga
-
1
logb b) loga-logb
c) loga+logb d)
1
loga
+
2
logb

64. DEFINITE INTEGRALS
10)
0
/𝟒
etanx
cos2x
dx =
a) e-1 b) e-1-1 c) e-1+1 d) e-2-1

65. DEFINITE INTEGRALS
11)
1
𝟐
cos(logx)
x
dx =
a) sin(log2) b) sin(log3) c) –sin(log2) d) cos(log2)

66. DEFINITE INTEGRALS
12)
0
𝟏
(1+e−x)dx =
a) -1 b) 2 c) 1+e-1 d) 2-e-1

67. DEFINITE INTEGRALS
13)
0
/𝟐
esin2x .sin2x dx =
a) e b) e+1 c) e-1 d) 2e+1

68. DEFINITE INTEGRALS
14)
0
/𝟒
sin9x
cos11x
dx =
a) 10 b) 5 c)
1
10
d)
1
5

69. DEFINITE INTEGRALS
15)
−1
𝟏
xex dx =
a)
2
e b)
e𝟐
2
c) 2e d)
2
e𝟐

70. DEFINITE INTEGRALS
Thank you…