1. The Substitution Method
Thursday, 5 December 13

2. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Thursday, 5 December 13

3. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
Integrate
Thursday, 5 December 13
∫ (1+ x )
3 5
2
x dx

4. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
Integrate
∫ (1+ x )
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
Thursday, 5 December 13

5. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
3
Thursday, 5 December 13

6. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
du
2
= 3x
dx
3
Thursday, 5 December 13

7. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
du
2
= 3x
dx
2
du = 3x dx
3
Thursday, 5 December 13

8. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
du
2
= 3x
dx
2
du = 3x dx
du
2
= x dx
3
3
Thursday, 5 December 13

9. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
∴∫ (1+ x 3 )5 x 2 dx
du
2
= 3x
dx
2
du = 3x dx
du
2
= x dx
3
3
Thursday, 5 December 13

10. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
∴∫ (1+ x 3 )5 x 2 dx
du
2
= 3x
dx
u
2
du = 3x dx
du
2
= x dx
3
3
Thursday, 5 December 13

11. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
∴∫ (1+ x 3 )5 x 2 dx
du
2
= 3x
dx
u
2
du
du = 3x dx
3
du
2
= x dx
3
3
Thursday, 5 December 13

12. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
5 du
3 5 2
u = 1+ x
∴∫ (1+ x ) x dx = ∫ u
3
du
2
= 3x
dx
u
2
du
du = 3x dx
3
du
2
= x dx
3
3
Thursday, 5 December 13

13. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
5 du
3 5 2
u = 1+ x
∴∫ (1+ x ) x dx = ∫ u
3
du
2
= 3x
1 5
= ∫ u du
dx
3
u
2
du
du = 3x dx
3
du
2
= x dx
3
3
Thursday, 5 December 13

14. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
5 du
3 5 2
u = 1+ x
∴∫ (1+ x ) x dx = ∫ u
3
du
2
= 3x
1 5
= ∫ u du
dx
3
u
6
2
du
1u
du = 3x dx
=
+c
3
3 6
du
2
= x dx
3
3
Thursday, 5 December 13

15. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
∴∫ (1+ x 3 )5 x 2 dx
du
2
= 3x
dx
u
2
du
du = 3x dx
3
du
2
= x dx
3
3
Thursday, 5 December 13
du
= ∫u
3
1 5
= ∫ u du
3
6
1u
=
+c
3 6
6
u
=
+c
18
5

16. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
∴∫ (1+ x 3 )5 x 2 dx
du
2
= 3x
dx
u
2
du
du = 3x dx
3
du
2
= x dx
3
3
Thursday, 5 December 13
du
= ∫u
3
1 5
= ∫ u du
3
6
1u
=
+c
3 6
6
u
=
+c
18
5
but u = 1+ x
3

17. The Substitution Method
This section describes the substitution method that changes the form
of an integrand, preferably to one that we can integrate more easily.
Example
∫ (1+ x )
Integrate
3 5
2
x dx
Solution:
2
2
3
We notice the derivative of (1+ x ) is 3x , which differs from x in the
3
integrand only by a factor of 3. So let u = 1+ x .
u = 1+ x
∴∫ (1+ x 3 )5 x 2 dx
du
2
= 3x
dx
u
2
du
du = 3x dx
3
du
2
= x dx
3
3
Thursday, 5 December 13
du
3
but u = 1+ x
= ∫u
3
3 6
(1+ x )
3 5 2
∴ ∫ (1+ x ) x dx =
+c
1 5
= ∫ u du
18
3
6
1u
=
+c
3 6
6
u
=
+c
18
5

18. Substitution in a Definite Integral
Thursday, 5 December 13

19. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Thursday, 5 December 13

20. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
0
Thursday, 5 December 13
2x + 1 dx using u = 2x + 1

21. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
0
Solution:
Thursday, 5 December 13
2x + 1 dx using u = 2x + 1

22. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
0
Solution:
u = 2x + 1
Thursday, 5 December 13
2x + 1 dx using u = 2x + 1

23. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
0
Solution:
u = 2x + 1
du
=2
dx
Thursday, 5 December 13
2x + 1 dx using u = 2x + 1

24. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
Thursday, 5 December 13
2x + 1 dx using u = 2x + 1

25. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
2x + 1 dx using u = 2x + 1

26. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0

27. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
u =1

28. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1

29. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1
u=9

30. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
4
∴∫ 2x + 1dx
0
u =1
u=9

31. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
4
∴∫ 2x + 1dx
0
u
u =1
u=9

32. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
4
∴∫ 2x + 1dx
0
u
du
2
u =1
u=9

33. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1
u=9
4
9
0
1
∴∫ 2x + 1dx = ∫
u
du
2
du
u
2

34. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1
u=9
4
9
0
1
du
u
2
∴∫ 2x + 1dx = ∫
9
u
du
2
1
2
1
= ∫ u du
21

35. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1
u=9
4
9
0
1
du
u
2
∴∫ 2x + 1dx = ∫
9
u
du
2
1
2
1
= ∫ u du
21
9
1 ⎡2 ⎤
= ⎢ u ⎥
2 ⎣3 ⎦1
3
2

36. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1
u=9
4
9
0
1
du
u
2
∴∫ 2x + 1dx = ∫
9
u
du
2
⎞
1⎛
= ⎜ 9 −1 ⎟
3⎝
⎠
3
2
1
2
1
= ∫ u du
21
9
1 ⎡2 ⎤
= ⎢ u ⎥
2 ⎣3 ⎦1
3
2
3
2

37. Substitution in a Definite Integral
In evaluating a definite integral by use of change of variable, there
must be a corresponding change in the limits of integration.
Example
Evaluate
4
∫
2x + 1 dx using u = 2x + 1
0
Solution:
u = 2x + 1
du
=2
dx
du = 2dx
du
= dx
2
Thursday, 5 December 13
when x = 0
x=4
u =1
u=9
4
9
0
1
du
u
2
∴∫ 2x + 1dx = ∫
9
u
du
2
⎞
1⎛
= ⎜ 9 −1 ⎟
3⎝
⎠
3
2
1
2
1
= ∫ u du
21
26
=
3
9
1 ⎡2 ⎤
= ⎢ u ⎥
2 ⎣3 ⎦1
3
2
3
2

38. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
Thursday, 5 December 13

39. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
∫ f (x)dx
Thursday, 5 December 13
with
∫ g(u)du

40. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
∫ f (x)dx
with
u(b)
b
and
∫
a
Thursday, 5 December 13
∫ g(u)du
f (x)dx
with
∫
u(a)
g(u)du

41. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
∫ f (x)dx
with
u(b)
b
and
∫
∫ g(u)du
f (x)dx
with
∫
g(u)du
u(a)
a
It is hoped that the problem of finding
∫ g(u)du
Thursday, 5 December 13

42. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
∫ f (x)dx
with
u(b)
b
and
∫
∫ g(u)du
f (x)dx
with
∫
g(u)du
u(a)
a
It is hoped that the problem of finding
∫ g(u)du
Thursday, 5 December 13
is easier to find than

43. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
∫ f (x)dx
with
u(b)
b
and
∫
∫ g(u)du
f (x)dx
with
∫
g(u)du
u(a)
a
It is hoped that the problem of finding
∫ g(u)du
Thursday, 5 December 13
is easier to find than
∫ f (x)dx

44. Summary
This section introduced the most commonly used integration
technique, ‘substitution’, which replaces
∫ f (x)dx
with
u(b)
b
and
∫
∫ g(u)du
f (x)dx
with
∫
g(u)du
u(a)
a
It is hoped that the problem of finding
∫ g(u)du
is easier to find than
if it isn’t, try another substitution.
Thursday, 5 December 13
∫ f (x)dx