Integration by Parts Objective: To integrate problems without a u-substitution.
§4.3 Integration by Substitution.
description
Transcript of §4.3 Integration by Substitution.
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The student will learn about:
§4.3 Integration by Substitution.
integration by substitution.
differentials, and
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Introduction
The use of the “Chain Rule” greatly expanded the range of functions that we could differentiate. We need a similar property in integration to allow us to integrate more functions. The method we are going to use is called integration by substitution.
But first lets practice some basic integrals.
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Preliminaries.
For a differentiable function f (x), the differential df is
df = f ′(x) dx This is the derivative.
This comes from previous notation:
)x('fdx
df
dx)x('fdf
Multiply both sides by dx to get the differential
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Examples.
Function
f (x) = x 3
Differential df
df = 3 x 2 dx
f (x) = ln x df = 1/x dx
f (x) = e 3x df = dxxe33
f (x) = x 2 - 3x + 5 df = (2x – 3) dx
We are taking a derivative now !!!
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Reversing the Chain Rule. Recall the chain rule:
This means that in order to integrate 5 (x 3 + 3) 4 its derivative, 3x 2, must be present. The “chain” must be present in order to integrate.
3 5 3 24d(x 3 ) 5(x 3 ) or
dx3x
23 5 3 4d (x 3 ) [5(x 3x3 ) ] dx A differential
3 5 3 4 2d (x 3 ) [5(x 3 ) 3x ] dx 3 5 3 4 2(x 3 ) c [5(x 3 ) 3x ] dx
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General Indefinite Integral Formulas.
∫ un du = 1n,C1n
u 1n
∫ e u du = e u + C
1lnu C
ud u
Note the chain “du” is present!
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Integration by Substitution
Note that the derivative of x 5 – 2 , (i.e. 5x 4, the chain), is present and the integral is in the power rule form ∫ u n du.
dx)x5()2x(:problemfollowingtheConsider 435
There is a nice method of integration called “integration by substitution” that will handle this problem.
This is the power rule form ∫ u n du.
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Example by Substitution
∫ u 3 du which is =
By example (Substitution): Find ∫ (x5 - 2)3 (5x 4) dx
For our substitution let u = x 5 - 2,
5x 4, and du = 5x 4 dxthen du/dx =
and the integral becomes
For our substitution let u = x 5 - 2,
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
∫ (x5 - 2)3 (5x 4) dx =
“Power Rule”
∫ un du = n 1u
Cn 1
Continued
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Example by Substitution
c4
u4
∫ u 3 du which is =
By example (Substitution): Find ∫ (x5 - 2)3 (5x 4) dx
u = x 5 - 2,
and reverse substitution yields
c
4
2x 45
Remember you may differentiate
to check your work!
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Integration by Substitution. Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
Step 2. Express the integrand entirely in terms of u and du, completely eliminating the original variable.
Step 3. Evaluate the new integral, if possible.
Step 4. Express the antiderivative found in step 3 in terms of the original variable. (Reverse the substitution.)
Step 1. Select a substitution that appears to simplify the integrand. In particular, try to select u so that du is a factor of the integrand.
Remember you may differentiate to check your work!
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Example∫ (x3 - 5)4 (3x2) dx
Step 1 – Select u.
Let u = x3 - 5 and then du =Step 2 – Express integral in terms of u.
∫ (x3 - 5)4 (3x2) dx = ∫ u4 du
Step 3 – Integrate.
∫ u4 du =
Step 4 – Express the answer in terms of x.
3x2 dx
c5
u5
c5
)5x(c
5
u 535
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
Did I mention you may differentiate to check your work?
“Power Rule”
∫ un du = n 1u
Cn 1
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Example∫ (x 2 + 5) 1/2 (2x ) dx
Step 1 – Select u.Let u = x2 + 5 and then du =
Step 2 – Express integral in terms of u.
∫ (x2 + 5) 1/2 (2x) dx = ∫ u 1/2 du
Step 3 – Integrate.
∫ u 1/2 du =
Step 4 – Express the answer in terms of x.
2x dx
c)5x(3
22
32
cu3
2c
23
u2
323
cu3
22
3
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
“Power Rule”
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Multiplying Inside and Outside by Constants
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Multiplying Inside and Outside by Constants
If the integral does not exactly match the form
∫un du,
we may sometimes still solve the integral by multiplying by constants.
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Substitution Technique – Example 1
1.
∫ (x3 - 5)4 (x2) dx =
∫ (x3 - 5)4 (x2) dx Let u =
∫ (x3 - 5)4 ( x2) dx =
In this problem we had to insert a multiple 3 in order to get things to work out. We did this by also dividing by 3 elsewhere.
x3 – 5 then du = 3x2 dx
Note – we need a 3.
In this problem we had to insert a multiple 3 in order to get things to work out. We did this by also dividing by 3 elsewhere. Caution – a constant can be adjusted but a variable cannot.
=c+5
u
3
1 5
c+15
u 5
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
Remember you may differentiate to check your work!
3 51(x 5) c
15
41u du
3
1
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“Power Rule”
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2. Let u = 4x3 then du = 12x2 dx
Note – we need a 12.
dxex3x42
32 4x112 x e dx
12
ce12
1 3x4
In this problem we had to insert a multiple 12 in order to get things to work out. We did this by also dividing by 12 elsewhere.
dxex3x42
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
Remember you may differentiate to check your work!
Substitution Technique – Example 2
u1e du
12 u1
e c12
“Exponential Rule” ∫ e u du = e u + c
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3. Let u = 5 – 2x2 then du = - 4x dx
Note – we need a - 4.
dx)x25(
x52
dx)x25(
x52
dx)x25(
x4
4
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In this problem we had to insert a multiple - 4 in order to get things to work out.
duu
1
4
15
duu4
1 5
c)x25(16
142
cu
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1 4-
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
Substitution Technique – Example 3
2 41(5 2x ) c
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“Power Rule”
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Applications §6-2, #68. The marginal price of a supply level of x bottles of baby shampoo per week is given by
Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle.
Continued on next slide.
2)25x3(
300)x('p
To find p (x) we need the ∫ p ‘ (x) dx
Step
1
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Applications - continuedThe marginal price of a supply level of x bottles of baby shampoo per week is given by
Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle.
Continued on next slide.
2)25x3(
300)x('p
Let u =
cu 1100
22
300dx 300 (3x 25) dx
(3x 25)
duu100 2
3x + 25
3 dx
Step
1
and du =
2 2303
3
0300 (3x 25) dx (3x 25) dx
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Applications - continuedThe marginal price of a supply level of x bottles of baby shampoo per week is given by
Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle.
Continued on next slide.
2)25x3(
300)x('p
With u = 3x + 25,
so1 100
p(x) 100(3x 25) c c3x 25
Remember you may differentiate to check your work!
Step
1
cu 1100 1100(3x 25) c
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Applications - continuedThe marginal price of a supply level of x bottles of baby shampoo per week is given by
Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle.
Continued on next slide.
2)25x3(
300)x('p
c25x3
100)x(p
Now we need to find c using the fact
That 75 bottles sell for $1.60 per bottle.
c25)75(3
10060.1
and c = 2
Step
2
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Applications - concludedThe marginal price of a supply level of x bottles of baby shampoo per week is given by
Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle.
2)25x3(
300)x('p
225x3
100)x(p
So
Step
2
This problem does not have a step 3!
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Summary.
2.
1. ∫ un du = 1n,C1n
u 1n
∫ e u du = e u + C
Culnduu
13.
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ASSIGNMENT
§4.3 on my website.
13, 14, 15, 16.