# Integration by Substitution  2018-11-07 8K 8 0 0

## Integration by Substitution - Description

AP Calculus. Mrs. . Mongold. Antidifferentiation. of a Composite Function. Let . g. be a function whose range is an interval . I. , and let . f. be a function that is continuous on . I. . If . g. ID: 720307 Download Presentation

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## Integration by Substitution

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### Presentations text content in Integration by Substitution

Slide1

Integration by Substitution

AP Calculus

Mrs.

Mongold

Slide2

Antidifferentiation of a Composite Function

Let

g

be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then If u = g(x), then du= g’(x)dx and

Slide3

Exploration

Recognizing Patterns

The integrand in each of the following integrals fits the pattern

f(g(x)g’(x). Identify the pattern and use the result to evaluate the integral.A)The next three integrals are similar to the first three. Show how you can multiply and divide by a constant to evaluate these integrals.

Slide4

Example

Evaluate

Letting

g(x) = , you obtain g’(x) = 2xAnd f(g(x)) = [g(x)]2So,

Slide5

Example

Evaluate

Letting

u = , you obtain du = 2x dxAnd f(g(x)) = [g(x)]2So,

Slide6

Example

Evaluate

Letting u =

, you obtain du = 2x dxNow substituteSo,

Slide7

Example

Evaluate

Letting

u= , you obtain du= 2x dxNow substituteSo,

Slide8

Example

Evaluate

Letting

u= , you obtain du= 2x dxNow substituteSo, =

Slide9

Example

Evaluate

Letting

u= , you obtain du= 2x dxNow substituteSo, = Re-substitute 

Slide10

Example

Evaluate

Letting

g(x) = 5x, you obtain g’(x) = 5 And f(g(x)) = cos[g(x)]So,

Slide11

Example

Evaluate

Letting

u = 5x, you obtain du= 5 And f(g(x)) = cos[g(x)]So,

Slide12

Example

Evaluate

Letting

u = 5x, you obtain du = 5 Now substituteSo,

Slide13

Example

Evaluate

Letting

u = 5x, you obtain du = 5 Now substituteSo,

Slide14

Example

Evaluate

Letting

u = 5x, you obtain du = 5 Now substituteSo,

Slide15

Example

Evaluate

Letting

u = 5x, you obtain du = 5 Now substituteSo, Re-substitute

Slide16

Example

Evaluate

Letting

u = 5x, you obtain du = 5 Now substituteSo, Re-substitute

Slide17

Example

Evaluate

Similar to Example 1, except the integrand is missing a factor of 2. Recognizing that 2x is the derivative of x

2 +1, you can let g(x) = x2+1 and supply the 2x as follows

Slide18

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx

Slide19

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx

Oh no, I don’t have a 2x dx!

What do I do?!?!

Slide20

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx

Oh no, I don’t have a 2x dx!

What do I do?!?!

No worries… solve for dx and

Do a second substitution

Slide21

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx and dx =

Slide22

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx and dx =

Slide23

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx and dx =

Slide24

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx and dx =

Slide25

Example

Evaluate

Let u = x

2 + 1, then du = 2x dx and dx =

Slide26

Examples

Let

u

= 2x -1 and du = 2dxNow using and dx =du/2 =

Slide27

Examples

Let

u

= 2x -1 and du = 2dxNow using and dx =du/2 =

Slide28

Examples

Let

u

= 2x -1 and du = 2dxNow using and dx =du/2 =

Slide29

Examples

Let

u

= 2x -1 and du = 2dxNow using and dx =du/2 =

Slide30

Examples

Let

u

= 2x -1 and du = 2dxNow using and dx =du/2 =

Slide31

Examples

Let

u

= 2x -1 and du = 2dxNow using and dx =du/2 =

Slide32

Examples

Let

u =

2x – 1, so x = (u +1)/2Now using substitution, you obtain the following

Slide33

Examples

Let

u =

2x – 1, so du =2 dxNow using substitution, you obtain the following

Slide34

Examples

Let

u =

2x – 1, so du =2 dxNow using substitution, you obtain the following

Hmm….is this a problem…

d

x = du/2 but I still have an x

Slide35

Examples

Let

u =

2x – 1, so du = 2dx dx = du/2 and x = (u +1)/2Now using substitution, you obtain the following

Slide36

Examples

Let

u =

2x – 1, so x = (u +1)/2Now using substitution, you obtain the following

Slide37

Examples

Let

u =

2x – 1, so x = (u +1)/2Now using substitution, you obtain the following

Slide38

Examples

Let

u =

2x – 1, so x = (u +1)/2Now using substitution, you obtain the following

Slide39

Example

Because sin

2

3x = (sin 3x)2 let u = sin 3x and du = 3cos 3x dx Substituting u and du/3 gives the following

Slide40

Example

Because sin

2

3x = (sin 3x)2 let u = sin 3x and du = 3cos 3x dx Substituting u and du/3 gives the following

Slide41

Example

Because sin

2

3x = (sin 3x)2 let u = sin 3x and du = 3cos 3x dx Substituting u and du/3 gives the following

Slide42

Example

Because sin

2

3x = (sin 3x)2 let u = sin 3x and du = 3cos 3x dx Substituting u and du/3 gives the following

Slide43

Example

Because sin

2

3x = (sin 3x)2 let u = sin 3x and du = 3cos 3x dx Substituting u and du/3 gives the following

Slide44

Example

Because sin

2

3x = (sin 3x)2 let u = sin 3x and du = 3cos 3x dx Substituting u and du/3 gives the following

Slide45

Guidelines for Substitution

1. Choose a substitution u = g(x). Usually, it is best to choose the “inner” part of a composite function, such as a quantity raised to a power.

2. Computer du = g’(x) dx

3. Rewrite the integral in terms of the variable u4. Evaluate the resulting integral in terms of u5. Replace u with g(x)to obtain an antiderivative in terms of x6. Check your answer by differentiating

Slide46

Remember General Power Rule

If g is a differentiable function of x, then

Equivalently if u = g(x), then

Slide47

Examples

Slide48

Examples

Slide49

Examples

Slide50

Examples

Slide51

Examples

Slide52

Examples

Slide53

Examples

Slide54

Examples

Slide55

Examples

Slide56

Examples

Slide57

Examples

Slide58

Examples

Slide59

Examples

Slide60

Examples

Slide61

Examples

Slide62

Substitution For Definite Integrals

If the function u = g(x) has a continuous derivative on the closed interval [a, b] and f is continuous on the range of g, then

Slide63

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide64

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide65

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide66

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide67

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide68

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide69

Examples

Let

u

= and du = 2x dxFind Lower Limit when x=0, u = 02 + 1 = 1Find Upper Bound when x = 1, u = 12 + 1 = 2Now Substitute =

=

Slide70

Example

Evaluate

Let

u = and u du = dxFind Lower Limit u = = 1Find Upper Limit u = = 3

Slide71

Example

Evaluate

Let

u = and u du = dxFind Lower Limit u = = 1Find Upper Limit u = = 3

Slide72

Example

Evaluate

Let

u = and u du = dxFind Lower Limit u = = 1Find Upper Limit u = = 3

Slide73

Example

Evaluate

Let

u = and u du = dxFind Lower Limit u = = 1Find Upper Limit u = = 3

Slide74

Example

Evaluate

Let

u = and u du = dxFind Lower Limit u = = 1Find Upper Limit u = = 3

Slide75

Slide76

Slide77

Slide78

Slide79

Integration of Even and Odd Functions

Let

f

be integrable on the closed interval [-a, a].1. If f is an even function, then 2. If f is an odd function, then

Slide80

How does this work?!?!

EVEN Function PROOF

- If f is even what do we know about it?

We know that f(x)=f(-x)Symmetry happensThink about the graph of a an even function about the origin

Slide81

How does this work?!?!

EVEN Function PROOF

- If f is even what do we know about it?

We know that f(x)=f(-x)Symmetry happensThink about the graph of a an even function about the origin

Slide82

How does this work?!?!

EVEN Function PROOF

- If f is even what do we know about it?

We know that f(x)=f(-x)Symmetry happensThink about the graph of a an even function about the origin

Slide83

How does this work?!?!

EVEN Function PROOF

- If f is even what do we know about it?

We know that f(x)=f(-x)Symmetry happensThink about the graph of a an even function about the origin

Slide84

How does this work?!?!

EVEN Function PROOF

- If f is even what do we know about it?

We know that f(x)=f(-x)Symmetry happensThink about the graph of a an even function about the origin

Slide85

How does this work?!?!

EVEN Function PROOF

- If f is even what do we know about it?

We know that f(x)=f(-x)Symmetry happensThink about the graph of a an even function about the origin

Slide86

ODD function PROOF

Slide87

Example

Evaluate

Letting

f(x) = sin3 xcosx + sinxcosx produces f(-x) = sin3(-x)cos(-x) + sin(-x)cos(-x) = -sin3x cosx – sinx cosx = - f(x)

So,

f

is an odd function and because [- , is symmetric we can conclude that = 0

Slide88

Example

Evaluate

Letting

f(x) = sin3 xcosx + sinxcosx produces f(-x) = sin3(-x)cos(-x) + sin(-x)cos(-x) = -sin3x cosx – sinx cosx = - f(x)

So,

f

is an odd function and because [- , is symmetric we can conclude that = 0

Slide89

Example

Evaluate

Letting

f(x) = sin3 xcosx + sinxcosx produces f(-x) = sin3(-x)cos(-x) + sin(-x)cos(-x) = -sin3x cosx – sinx cosx = - f(x)

So,

f

is an odd function and because [- , is symmetric we can conclude that = 0

Slide90

Example

Evaluate

Letting

f(x) = sin3 xcosx + sinxcosx produces f(-x) = sin3(-x)cos(-x) + sin(-x)cos(-x) = -sin3x cosx – sinx cosx = - f(x)

So,

f

is an odd function and because [- , is symmetric we can conclude that = 0

Slide91

Example

Evaluate

Letting

f(x) = sin3 xcosx + sinxcosx produces f(-x) = sin3(-x)cos(-x) + sin(-x)cos(-x) = -sin3x cosx – sinx cosx = - f(x)

So,

f

is an odd function and because [- , is symmetric we can conclude that = 0

Slide92

Example

Evaluate

Letting

f(x) = sin3 xcosx + sinxcosx produces f(-x) = sin3(-x)cos(-x) + sin(-x)cos(-x) = -sin3x cosx – sinx cosx = - f(x)

So,

f

is an odd function and because [- , is symmetric we can conclude that = 0

Slide93

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