11

The student will learn about:

§4.3 Integration by Substitution.

integration by substitution.

differentials, and

22

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.

33

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

44

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!

1010

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

1212

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

14

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.

1515

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

16

“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

2020

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

2121

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

2222

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.