§4.3 Integration by Substitution.
The student will learn about:
differentials, and
integration by
substitution.
1
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.
2
Preliminaries.
For a differentiable function f (x), the
differential df is
df = f ′(x) dx
This is the derivative.
This comes from previous notation:
df
 f' ( x )
dx
Multiply both sides by dx to get the differential
df  f ' ( x )  dx
3
Examples.
Function
f (x) = x 3
Differential df
df = 3 x 2 dx
We are taking a derivative now !!!
f (x) = x 2 - 3x + 5
df = (2x – 3) dx
f (x) = ln x
df = 1/x dx
f (x) =
e 3x
3x
df = 3 e dx
4
Reversing the Chain Rule.
Recall the chain rule:
d
(x3  3 )5  5(x3  3 )4 3x2 or
dx
d (x3  3 )5  [ 5(x3  3 )4 3x2 ] dx
A differential
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
4
2
d
(x

3)

[
5(x

3)
3x
] dx


(x3  3)5  c   [ 5(x3  3)4 3x2 ] dx
5
General Indefinite Integral Formulas.
n 1
∫
un
u
du =
 C, n   1
n 1
∫ e u du = e u + C

1
du  ln u  C
u
Note the chain “du” is present!
6
Integration by Substitution
Consider the following problem :
5
3
4
(
x

2
)
(
5
x
) dx

This is the power rule form ∫ u n du.
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.
There is a nice method of integration called
“integration by substitution” that will handle this
problem.
7
Example by Substitution
∫ un du =
un  1
C
n1
By example (Substitution): Find ∫ (x5 - 2)3 (5x 4) dx
For our substitution let u = x 5 - 2,
then du/dx = 5x 4, and du = 5x 4 dx
and the integral becomes
Is it “Power Rule”,
Rule”
“Exponential Rule”
or the “Log Rule”?
∫ (x5 - 2)3 (5x 4) dx = ∫ u 3 du which is =
Continued
8
Example by Substitution
By example (Substitution): Find ∫ (x5 - 2)3 (5x 4) dx
u = x 5 - 2,
4
∫
u3
du
u
c
which is =
4
and reverse substitution yields
x
5
 2
4
4
 c
Remember you may differentiate
to check your work!
9
Integration by Substitution.
Is it “Power Rule”, “Exponential Rule” or the “Log Rule”?
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
Remember you may differentiate to check your work!
un  1
∫ un du = n  1  C
Is it “Power
“Power
Rule”Rule”, “Exponential
Rule” or the “Log Rule”?
Example
∫ (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
Did I mention you may
Step 3 – Integrate.
5
differentiate to check your wo
u
∫ u4 du =
c
5
Step 4 – Express the answer in terms of x.
u5
( x 3  5) 5
c
c
5
5
11
Example
∫
(x 2 +
5) 1/2
(2x ) dx
Is it “Power Rule”, “Exponential
Rule” or the “Log Rule”?
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.
3
u
1/2
∫ u du = 3
2
2
“Power Rule”
2 32
c u c
3
Step 4 – Express the answer in terms of x.
2 32
u c
3
3
2 2
(x  5) 2  c
3
12
Multiplying Inside and Outside
by Constants
13
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.
14
Substitution Technique – Example 1
Is it “Power
“Power
Rule”,
Rule”“Exponential Rule” or the “Log Rule”?
1. ∫ (x3 - 5)4 (x2) dx Let u = x3 – 5 then du = 3x2 dx
∫
(x3
-
5)4
1
dx =
∫ (x3 - 5)4 ( 3 x2) dx =
3 5
5
(x2)
1 u
u
1 4
+c=
+c
u du 

15
3 5
3
Note – we
need a 3.
1 3
(x  5)5  c
15
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.
15
Remember you may differentiate to check your work!
Substitution Technique – Example 2
Is it “Power Rule”,
“Exponential Rule”
Rule” or
“Exponential
∫ etheu “Log
du =Rule”?
eu +
2.
2 4x 3
x e
2 4x 3
x e
c
dx Let u = 4x3 then du = 12x2 dx
dx 
1
2 4x 3
12 x e dx 

12
Note – we
1 u
1 u
1 4x 3
need a 12.
e du 
e c
e c

12
12
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.
Remember you may differentiate to check your work!
16
Substitution Technique – Example 3
3. 
x
( 5  2 x 2 )5
x
dx
Let u = 5 – 2x2 then du = - 4x dx
1
4x
dx   
dx 

2 5
2
5
4 ( 5  2x )
( 5  2x )
Note – we
need a - 4.
1 5
1 u -4
1 1
c
  5 du    u du 
4
4
4
4 u
1
(5  2x 2 )  4  c 
16
1
c
2 4
16 (5  2x )
In this problem we had to insert a multiple - 4 in
order to get things to work out.
Is “Power
it “Power
Rule”
Rule”, “Exponential Rule” or the “Log Rule”?
17
Applications
To find p (x) we need the ∫ p ‘ (x) dx
Step 1
§6-2, #68. The marginal price of a supply level of x
bottles of baby shampoo per week is given by
300
p' ( x ) 
( 3x  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.
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 ) 
( 3x  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.
 (3x  25)2
dx  300  (3x  25)
300  (3x  25)
2
2
dx
Let u = 3x + 25
and du = 3 dx
Step 1
300
300
2
2
dx 
(3x

25)
3
dx

100
u
du 


3
 100 u 1  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 ) 
( 3x  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 (3x  25)
1
Step 1
With u = 3x + 25,  100 u  1  c   100(3x  25) 1  c
100
c
c
3x  25
Remember you may differentiate to check your work!
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 ) 
( 3x  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.
Now we need to find c using the fact
Step 2
 100
p ( x) 
c
3x  25
That 75 bottles sell for $1.60 per bottle.
 100
1.60 
c
3(75)  25
and c = 2
Continued on next slide.
21
Applications - concluded
The marginal price of a supply level of x bottles of baby
shampoo per week is given by
300
p' ( x ) 
( 3x  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.
 100
p ( x) 
2
3x  25
Step 2
So
This problem does not have a step 3!
22
Summary.
n 1
un
u
 C, n   1
du =
n 1
1.
∫
2.
∫ e u du = e u + C
3.
1
 u du  ln u  C
23
ASSIGNMENT
§4.3 on my website.
13, 14, 15, 16.
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