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.
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§12.2 Integration by Substitution.
The student will learn about:
integration by substitution,
some substitution techniques, and
applications.
2
Reversing the Chain Rule.
Recall the chain rule:
d
f [ g ( x ) ]  f '[ g ( x ) ] g ' ( x )
dx
This implies that:
 f ' [ g ( x ) ] g ' ( x ) dx  f [ g ( x ) ]  C
This means that in order to integrate g (x) its
derivative, g ‘ (x), must be present.
3
General Integral Formulas
[ f ( x )]
1.
∫ [f (x) ] n f ‘(x) dx =
2.
∫ e f (x) f ‘(x) dx = e f (x) + C
3.

1
n1
n1
 C, n   1
f ' ( x ) dx  ln f ( x )  C
f (x)
4
Example
5
3
4
 ( x  2 ) ( 5 x ) dx
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 .
∫ [f (x) ] f ‘(x) dx =
n
It follows that:
5
3
4
 ( x  2 ) ( 5 x ) dx 
[ f ( x )]
n1
 C, n   1
n1
5
( x 2 )
4
 C
4
5
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
6
Examples
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
3. If y = f (x) = ln (3x -5), then
dy = f ‘ (x) dx =
3
3x  5
dx
7
Integration by Substitution
By example (Substitution): Find ∫ (x2 + 1)5 (2x) dx
For our substitution let u = x2 + 1,
then du/dx = 2x, and du = 2x dx
and the integral becomes
∫ u5 du which is = u6/6 + C
and reverse substitution yields
(x2 + 1)6/6 + C.
8
General Indefinite Integral Formulas.
∫
un
du =
u
n1
n1
 C, n   1
∫ e u du = e u + C

1
du  ln u  C
u
9
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.)
10
Example 2
∫ (x3 - 5)4 (3x2) dx
Step 1 – Select u.
Let u = x3 - 5 and then du = 3x2 dx
Step 2 – Express integral in terms of u.
∫ (x3 - 5)4 (3x2) dx = ∫ u4 du
Step 3 – Integrate.
∫ u4 du = u5/5 + c
Step 4 – Express the answer in terms of x.
u5/5 + c = (x3 – 5) 5/5 + c
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Example 3
∫ (x 2 + 5) 1/2 (2x ) dx
Step 1 – Select u.
Let u = x2 + 5 and then du = 2x dx
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 + c
Step 4 – Express the answer in terms of x.
2/3 u 3/2 + c = 2/3 (x2 + 5) 3/2 + c
12
Substitution Technique 1 – Example 1
1. ∫ (x3 - 5)4 (x2) dx
Let u = x3 – 5 then du = 3x2 dx
∫ (x3 - 5)4 (x2) dx = 1/3 ∫ (x3 - 5)4 (3x2) dx =
1/3 ∫ u4 du = 1/3 u5/5 = 1/15 u5 =
1/15 (x3 – 5)5 + c
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.
13
Substitution Technique 1 – Example 2
2
4x
3
2
4x
3
2.  x e
x e
1/12 ∫
dx
eu
dx 
Let u = 4x3 then du = 12x2 dx
1
12
du = 1/12
2
 12 x e
eu
4x
+c =
3
dx 
1
12
e
4x
3
 c
Note – we
need a 12.
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.
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Substitution Technique 1 – Example 3
3.
x


2 5
( 5  2x )
x
2 5
( 5  2x )

1

1
4 u5
dx
dx 
du  
1/16 (5 –
Let u = 5 – 2x2 then du = - 4x dx
2x2) - 4

1
4
1

4x
2 5
4 ( 5  2x )
u
5
+ c =
dx 
Note – we
need a - 4.
du  -1/4 · u - 4 /-4 =
1
16 ( 5  2 x )
2
4
 c
In this problem we had to insert a multiple - 4 in
order to get things to work out.
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Substitution Technique 2 – Example 1
4.  x  x  6  dx
8
Let u = x + 6 then du = dx
∫ x · (x + 6)8 dx = ∫ x u8 du = and this is ok, but
we need to get rid of the x. However, since u = x + 6,
x = u – 6, so the integral becomes
∫ (u – 6) · u8 du = ∫ u9 – 6u8 du =
u10/10 – 6 u9/9 + c = 1/10 · (x + 6)10 – 2/3 · (x + 6)9 +
c
In this problem we had to be somewhat creative
because of the extra x.
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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
p'(x ) 
300
( 3 x  25 )
2
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.
To find p (x) we need the ∫ p ‘ (x) dx
Continued on next slide.
17
Applications - continued
The marginal price of a supply level of x bottles of baby
shampoo per week is given by
300
p'(x ) 
( 3 x  25 )
2
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.

300
( 3 x  25 )
100 
dx  . . .
3
( 3 x  25 )
 100
u
2
 c
2
Let u = 3x + 25 and du = 3 dx
dx  100 
1
u
2
du  100  u  2 du 
Continued on next slide.
18
Applications - continued
The marginal price of a supply level of x bottles of baby
shampoo per week is given by
300
p'(x ) 
( 3 x  25 )
2
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.
With u = 3x + 25,
 100
 c
becomes
u
p (x) 
 100
3 x  25
 c
Continued on next slide.
19
Applications - continued
The marginal price of a supply level of x bottles of baby
shampoo per week is given by
300
p'(x ) 
( 3 x  25 )
2
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.
p (x) 
 100
3 x  25
 c
Now we need to find c using the fact
That 75 bottles sell for $1.60 per bottle.
1 . 60 
 100
3 ( 75 )  25
 c
and c = 2
Continued on next slide.
20
Applications - continued
The marginal price of a supply level of x bottles of baby
shampoo per week is given by
300
p'(x ) 
( 3 x  25 )
2
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.
So
p (x) 
 100
3 x  25
 2
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Summary.
u
 C, n   1
1.
∫
2.
∫ e u du = e u + C
3.

un
n1
1
du =
n1
du  ln u  C
u
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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.
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