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