indefinite and definite integrals using the substitution method
TRANSCRIPT
The Substitution Method
Thursday, 5 December 13
The Substitution MethodThis 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
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dx
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dxThursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u
∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
= u5 du3∫∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
= u5 du3∫
= 13
u5 du∫
∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
= u5 du3∫
= 13
u5 du∫= 13u6
6+ c
∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
= u5 du3∫
= 13
u5 du∫= 13u6
6+ c
= u6
18+ c
∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
= u5 du3∫
= 13
u5 du∫= 13u6
6+ c
= u6
18+ c
but u = 1+ x3∴
Thursday, 5 December 13
The Substitution MethodThis section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily.
Example
Integrate (1+ x3)5 x2 dx∫Solution: We notice the derivative of is , which differs from in the integrand only by a factor of 3. So let
(1+ x3) 3x2 x2u = 1+ x3.
u = 1+ x3dudx
= 3x2
du = 3x2dxdu3
= x2dx
(1+ x3)5 x2 dx∫
u du3
= u5 du3∫
= 13
u5 du∫= 13u6
6+ c
= u6
18+ c
but u = 1+ x3
∴ (1+ x3)5 x2 dx∫ = (1+ x3)6
18+ c
∴
Thursday, 5 December 13
Substitution in a Definite Integral
Thursday, 5 December 13
Substitution in a Definite IntegralIn 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
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫ u = 2x +1
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution:
u = 2x +1
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
u = 2x +1
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
u = 2x +1
dudx
= 2
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
u = 2x +1
dudx
= 2
du = 2dx
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
u = 2x +1
dudx
= 2
du = 2dx
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
u = 2x +1
dudx
= 2
du = 2dx
when x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
u = 2x +1
dudx
= 2
du = 2dx
when u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
u = 2x +1
dudx
= 2
du = 2dx
whenx = 4
u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u
∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u du2
∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u du2
= u du21
9
∫∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u du2
= u du21
9
∫
= 12
u12
1
9
∫ du
∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u du2
= u du21
9
∫
= 12
u12
1
9
∫ du
∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
= 1223u32⎡
⎣⎢
⎤
⎦⎥
1
9
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u du2
= u du21
9
∫
= 12
u12
1
9
∫ du
= 13932 −1
32
⎛⎝⎜
⎞⎠⎟
∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
= 1223u32⎡
⎣⎢
⎤
⎦⎥
1
9
Thursday, 5 December 13
Substitution in a Definite IntegralIn evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration.Example
Evaluate using 2x +1dx0
4
∫Solution: u = 2x +1
du2
= dx
2x +10
4
∫ dx
u du2
= u du21
9
∫
= 12
u12
1
9
∫ du
= 13932 −1
32
⎛⎝⎜
⎞⎠⎟
∴
u = 2x +1
dudx
= 2
du = 2dx
whenu = 9x = 4u = 1x = 0
= 1223u32⎡
⎣⎢
⎤
⎦⎥
1
9
= 263
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
f (x)dx∫ g(u)du∫with
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
f (x)dxa
b
∫
f (x)dx∫ g(u)du∫
g(u)duu(a)
u(b)
∫
with
withand
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
f (x)dxa
b
∫
f (x)dx∫ g(u)du∫
g(u)duu(a)
u(b)
∫
with
withand
It is hoped that the problem of finding
g(u)du∫
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
f (x)dxa
b
∫
f (x)dx∫ g(u)du∫
g(u)duu(a)
u(b)
∫
with
withand
It is hoped that the problem of finding
g(u)du∫ is easier to find than
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
f (x)dxa
b
∫
f (x)dx∫ g(u)du∫
g(u)duu(a)
u(b)
∫
with
withand
It is hoped that the problem of finding
g(u)du∫ f (x)dx∫is easier to find than
Thursday, 5 December 13
SummaryThis section introduced the most commonly used integration technique, ‘substitution’, which replaces
f (x)dxa
b
∫
f (x)dx∫ g(u)du∫
g(u)duu(a)
u(b)
∫
with
withand
It is hoped that the problem of finding
g(u)du∫ f (x)dx∫is easier to find than
if it isn’t, try another substitution.
Thursday, 5 December 13