1 warm up exercise for an average person, the rate of change of weight w with respect to height h is...

1

WARM UP EXERCISE

For an average person, the rate of change of weight W with respect to height h is given approximately by

dW/dh = 0.0015 h2

Find W (h), if W (60) = 108 pounds. Also, find the weight of a person who is 70 inches tall.

2

§12.2 Integration by Substitution.

The student will learn about:

integration by substitution, some substitution techniques, and

applications.

3

Reversing the Chain Rule.

Recall the chain rule:

)x('g])x(g['f])x(g[fdx

d

This implies that:

C])x(g[fdx)x('g])x(g['f

This means that in order to integrate g (x) its derivative, g ‘ (x), must be present.

4

General Integral Formulas

1. ∫ [f (x) ] n f ‘(x) dx = 1n,C1n

)]x(f[ 1n

2. ∫ e f (x) f ‘(x) dx = e f (x) + C

C)x(flndx)x('f)x(f

13.

5

Example

Note that the derivative of x 5 – 2 , (i.e. 5x 4 ), is present and the integral appears to be in the chain rule form ∫ [f (x) ] n f ‘(x) dx, with f (x) = x 5 – 2 .

dx)x5()2x( 435

It follows that:

dx)x5()2x( 435C

4

)2x( 45

∫ [f (x) ] n f ‘(x) dx = 1n,C1n

)]x(f[ 1n

6

Differential

If y = f (x) is a differentiable function, then

1. The differential dx of the independent variable x is any arbitrary real number.

2. The differential dy of the dependent variable y is defined as the product of f ‘ (x) and dx – that is, as

dy = f ‘ (x) dx

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Examples

1. If y = f (x) = x 5 – 2 , then

dy = f ‘ (x) dx =

2. If y = f (x) = e 5x , then

dy = f ‘ (x) dx =

3. If y = f (x) = ln (3x -5), then

dy = f ‘ (x) dx = dx5x3

3

1. If y = f (x) = x 5 – 2 , then

dy = f ‘ (x) dx = 5x 4 dx

2. If y = f (x) = e 5x , then

dy = f ‘ (x) dx = 5e 5x dx

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Integration by Substitution

∫ u5 du which is =

By example (Substitution): Find ∫ (x2 + 1)5 (2x) dx

For our substitution let u = x2 + 1,

2x, and du = 2x dx

u6/6 + C

and reverse substitution yields

(x2 + 1)6/6 + C.

then du/dx = and the integral becomes

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General Indefinite Integral Formulas.

∫ un du = 1n,C1n

u 1n

∫ e u du = e u + C

Culnduu

1

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Integration by 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.

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.)

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Example 2

∫ (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 = u5/5 + cStep 4 – Express the answer in terms of x.

u5/5 + c = (x3 – 5) 5/5 + c

3x2 dx

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Example 3

∫ (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 = u 3/2/(3/2) + c = 2/3 u 3/2 + cStep 4 – Express the answer in terms of x.

2/3 u 3/2 + c = 2/3 (x2 + 5) 3/2 + c

2x dx

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Substitution Technique 1 – Example 1

1.

∫ (x3 - 5)4 (x2) dx =

∫ (x3 - 5)4 (x2) dx Let u =

1/3 ∫ (x3 - 5)4 (3x2) 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

1/3 ∫ u4 du = 1/3 u5/5 = 1/15 u5 =

1/15 (x3 – 5)5 + cNote – 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.

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2. Let u = 4x3 then du = 12x2 dx

1/12 ∫ eu du = 1/12 eu + c = Note – we need a 12.

dxex3x42

dxex1212

1 3x42

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

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3. Let u =

-1/4 · u - 4 /-4 =

5 – 2x2 then du = - 4x dx

1/16 (5 – 2x2) - 4 + c =

Note – we need a - 4.

dx)x25(

x52

dx)x25(

x52

dx)x25(

x4

4

152

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

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4. Let u =

∫ x · (x + 6)8 dx =

∫ (u – 6) · u8 du =

u10/10 – 6 u9/9 + c =

x + 6 then du = dx

we need to get rid of the x. However, since u = x + 6,

x =

In this problem we had to be somewhat creative because of the extra x.

dx6xx 8

and this is ok, but∫ x u8 du =

u – 6, so the integral becomes

∫ u9 – 6u8 du =

1/10 · (x + 6)10 – 2/3 · (x + 6)9 + c

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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

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Applications - continued

The marginal price of a supply level of x bottles of baby shampoo per week is given by



2)25x3(

300)x('p

Let u =

cu

100

...dx)25x3(

3002

dx)25x3(

3100

2 duu

1100

2 duu100 2

3x + 25 and du = 3 dx

19





2)25x3(

300)x('p

With u = 3x + 25, cu

100

becomes

c25x3

100)x(p

20





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

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2)25x3(

300)x('p

225x3

100)x(p

So

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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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Practice Problems

§6.2; 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 43, 47, 51, 55, 59, 63, 67, 69, 71, 77, 79.

1 warm up exercise for an average person, the rate of change of weight w with respect to height h is...

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