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
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Slide1
Integration by Substitution
AP Calculus
Mrs.
MongoldSlide2
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.
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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
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Example
Evaluate
Let u = x
2 + 1, then du = 2x dx
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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
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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 xSlide35
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 differentiatingSlide46
Remember General Power Rule
If g is a differentiable function of x, then
Equivalently if u = g(x), then
Slide47
Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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Examples
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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
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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 originSlide81
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 originSlide82
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 originSlide83
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 originSlide84
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 originSlide85
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 originSlide86
ODD function PROOFSlide87
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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