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Transcript of INTEGRALS 5. 5.5 The Substitution Rule In this section, we will learn: To substitute a new variable...
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INTEGRALSINTEGRALS
5
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5.5The Substitution Rule
In this section, we will learn:
To substitute a new variable in place of an existing
expression in a function, making integration easier.
INTEGRALS
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Due to the Fundamental Theorem
of Calculus (FTC), it’s important to be
able to find antiderivatives.
INTRODUCTION
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However, our antidifferentiation formulas
don’t tell us how to evaluate integrals such
as22 1x x dx
Equation 1INTRODUCTION
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To find this integral, we use the problem-
solving strategy of introducing something
extra.
The ‘something extra’ is a new variable.
We change from the variable x to a new variable u.
INTRODUCTION
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Suppose we let u be the quantity under
the root sign in Equation 1, u = 1 + x2.
Then, the differential of u is du = 2x dx.
INTRODUCTION
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Notice that, if the dx in the notation for
an integral were to be interpreted as
a differential, then the differential 2x dx
would occur in Equation 1.
INTRODUCTION
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So, formally, without justifying our calculation,
we could write:
2 2
3/ 223
2 3/ 223
2 1 1 2
( 1)
x x dx x x dx
udu
u C
x C
Equation 2INTRODUCTION
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However, now we can check that we have
the correct answer by using the Chain Rule
to differentiate the final function of Equation 2:
2 3 2 2 1 232 23 3 2
2
( 1) ( 1) 2
2 1
dx C x x
dx
x x
INTRODUCTION
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In general, this method works whenever
we have an integral that we can write in
the form
∫ f(g(x))g’(x) dx
INTRODUCTION
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Observe that, if F’ = f, then
∫ F’(g(x))g’(x) dx = F(g(x)) + C
because, by the Chain Rule,
Equation 3INTRODUCTION
( ( )) '( ( )) '( )dF g x F g x g x
dx
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If we make the ‘change of variable’ or
‘substitution’ u = g(x), from Equation 3,
we have: '( ( )) '( ) ( ( ))
( )
'( )
F g x g x dx F g x C
F u C
F u du
INTRODUCTION
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Writing F’ = f, we get:
∫ f(g(x))g’(x) dx = ∫ f(u) du
Thus, we have proved the following rule.
INTRODUCTION
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SUBSTITUTION RULE
If u = g(x) is a differentiable function whose
range is an interval I and f is continuous
on I, then
∫ f(g(x))g’(x) dx = ∫ f(u) du
Equation 4
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SUBSTITUTION RULE
Notice that the Substitution Rule for
integration was proved using the Chain Rule
for differentiation.
Notice also that, if u = g(x), then du = g’(x) dx.
So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.
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SUBSTITUTION RULE
Thus, the Substitution Rule says:
It is permissible to operate with
dx and du after integral signs as if
they were differentials.
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SUBSTITUTION RULE
Find ∫ x3 cos(x4 + 2) dx
We make the substitution u = x4 + 2.
This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.
Example 1
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SUBSTITUTION RULE
Thus, using x3 dx = du/4 and the Substitution
Rule, we have:
Notice that, at the final stage, we had to return to the original variable x.
3 4 1 14 4
14
414
cos( 2) cos cos
sin
sin( 2)
x x dx u du u du
u C
x C
Example 1
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SUBSTITUTION RULE
The idea behind the Substitution Rule is to
replace a relatively complicated integral by
a simpler integral.
This is accomplished by changing from the original variable x to a new variable u that is a function of x.
Thus, in Example 1, we replaced the integral ∫ x3cos(x4 + 2) dx by the simpler integral ¼ ∫ cos u du.
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SUBSTITUTION RULE
The main challenge in using the rule is
to think of an appropriate substitution.
You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor.
This was the case in Example 1.
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SUBSTITUTION RULE
If that is not possible, try choosing u to be
some complicated part of the integrand—
perhaps the inner function in a composite
function.
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Finding the right substitution is
a bit of an art.
It’s not unusual to guess wrong.
If your first guess doesn’t work, try another substitution.
SUBSTITUTION RULE
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SUBSTITUTION RULE
Evaluate
Let u = 2x + 1.
Then, du = 2 dx.
So, dx = du/2.
2 1x dxE. g. 2—Solution 1
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Thus, the rule gives:
1 212
3 212
3 213
3 213
2 12
3/ 2
(2 1)
dux dx u
u du
uC
u C
x C
SUBSTITUTION RULE E. g. 2—Solution 1
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Another possible substitution is
Then,
So,
Alternatively, observe that u2 = 2x + 1. So, 2u du = 2 dx.
SUBSTITUTION RULE E. g. 2—Solution 2
2 1u x
2 1
dxdu
x
2 1dx x
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Thus,
SUBSTITUTION RULE E. g. 2—Solution 2
2
3
3 213
2 1
3
(2 1)
x dx u u du
u du
uC
x C
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SUBSTITUTION RULE
Find
Let u = 1 – 4x2. Then, du = -8x dx. So, x dx = -1/8 du and
21 4
xdx
x
1 21 18 82
21 18 4
1
1 4
(2 ) 1 4
xdx du u du
ux
u C x C
Example 3
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SUBSTITUTION RULE
The answer to the example could be
checked by differentiation.
Instead, let’s check it with a graph.
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SUBSTITUTION RULE
Here, we have used a computer to graph
both the integrand
and its indefinite integral
We take the case C = 0.
2( ) / 1 4f x x x 21
4( ) 1 4g x x
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SUBSTITUTION RULE
Notice that g(x):
Decreases when f(x) is negative Increases when f(x) is positive Has its minimum value when f(x) = 0
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So, it seems reasonable, from the graphical
evidence, that g is an antiderivative of f.
SUBSTITUTION RULE
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SUBSTITUTION RULE
Calculate ∫ e5x dx
If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore, 5 1
5
15
515
x u
u
x
e dx e du
e C
e C
Example 4
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SUBSTITUTION RULE
Find
An appropriate substitution becomes more obvious if we factor x5 as x4 . x.
Let u = 1 + x2.
Then, du = 2x dx.
So, x dx = du/2.
2 51 x x dxExample 5
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SUBSTITUTION RULE
Also, x2 = u – 1; so, x4 = (u – 1)2:
2 5 2 4 2
212
5/ 2 3/ 2 1/ 212
7 / 2 5/ 2 3/ 21 2 2 22 7 5 3
2 7 / 2 2 5/ 21 27 5
2 3/ 213
1 1 ( 1)2
( 2 1)
( 2 )
( 2 )
(1 ) (1 )
(1 )
dux x dx x x x dx u u
u u u du
u u u du
u u u C
x x
x C
Example 5
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SUBSTITUTION RULE
Calculate ∫ tan x dx
First, we write tangent in terms of sine and cosine:
This suggests that we should substitute u = cos x, since then du = – sin x dx, and so sin x dx = – du:
sintan
cos
xx dx dx
x
sintan
cosln | |
ln | cos |
x dux dx dx
x uu C
x C
Example 6
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SUBSTITUTION RULE
Since –ln |cos x| = ln(|cos x|-1)
= ln(1/|cos x|)
= ln|sec x|,
the result of the example can also be written
as ∫ tan x dx = ln |sec x| + C
Equation 5
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DEFINITE INTEGRALS
When evaluating a definite integral
by substitution, two methods are
possible.
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One method is to evaluate the indefinite
integral first and then use the FTC.
For instance, using the result of Example 2, we have:
DEFINITE INTEGRALS
44
0 0
43 213 0
3 2 3 21 13 3
2613 3
2 1 2 1
(2 1)
(9) (1)
(27 1)
x dx x dx
x
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DEFINITE INTEGRALS
Another method, which is usually preferable,
is to change the limits of integration when
the variable is changed.
Thus, we have the substitution rule for
definite integrals.
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If g’ is continuous on [a, b] and f
is continuous on the range of u = g(x),
then
( )
( )( ( )) '( ) ( )
b g b
a g af g x g x dx f u du
SUB. RULE FOR DEF. INTEGRALS Equation 6
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Let F be an antiderivative of f.
Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x).
So, by Part 2 of the FTC (FTC2), we have:
( ( )) '( ) ( ( ))
( ( )) ( ( ))
b b
aaf g x g x dx F g x
F g b F g a
SUB. RULE FOR DEF. INTEGRALS Proof
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However, applying the FTC2 a second time, we also have:
( ) ( )
( )( )( ) ( )
( ( )) ( ( ))
g b g b
g ag af u du F u
F g b F g a
SUB. RULE FOR DEF. INTEGRALS Proof
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Evaluate using Equation 6.
Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2
4
02 1x dx
Example 7SUB. RULE FOR DEF. INTEGRALS
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To find the new limits of integration, we
note that: When x = 0, u = 2(0) + 1 = 1 When x = 4, u = 2(4) + 1 = 9
Example 7SUB. RULE FOR DEF. INTEGRALS
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Thus,4 9
120 1
93 21 22 3 1
3 2 3 213
263
2 1
(9 1 )
x dx u du
u
Example 7SUB. RULE FOR DEF. INTEGRALS
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Observe that, when using Equation 6,
we do not return to the variable x after
integrating.
We simply evaluate the expression in u between the appropriate values of u.
SUB. RULE FOR DEF. INTEGRALS Example 7
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Evaluate
Let u = 3 - 5x.
Then, du = – 5 dx, so dx = – du/5.
When x = 1, u = – 2, and when x = 2, u = – 7.
2
21 (3 5 )
dx
xExample 8SUB. RULE FOR DEF. INTEGRALS
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Thus,2 7
2 21 2
7
5
7
2
1
(3 5 ) 5
1 1
5
1
5
1 1 1 1
5 7 2 14
dx du
x u
u
u
Example 8SUB. RULE FOR DEF. INTEGRALS
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Calculate
We let u = ln x because its differential du = dx/x occurs in the integral.
When x = 1, u = ln 1, and when x = e, u = ln e = 1.
Thus,
1
lne xdx
xExample 9SUB. RULE FOR DEF. INTEGRALS
121
1 00
ln 1
2 2
e x udx u du
x
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As the function f(x) = (ln x)/x in the example
is positive for x > 1, the integral represents
the area of the shaded region in this figure.
SUB. RULE FOR DEF. INTEGRALS Example 9
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SYMMETRY
The next theorem uses the Substitution
Rule for Definite Integrals to simplify
the calculation of integrals of functions that
possess symmetry properties.
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INTEGS. OF SYMM. FUNCTIONS
Suppose f is continuous on [–a , a].
a.If f is even, [f(–x) = f(x)], then
b.If f is odd, [f(-x) = -f(x)], then
0( ) 2 ( )
a a
af x dx f x dx
( ) 0a
af x dx
Theorem 7
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We split the integral in two:
0
0
0 0
( ) ( ) ( )
( ) ( )
a a
a a
a a
f x dx f x dx f x dx
f x dx f x dx
Proof—Equation 8INTEGS. OF SYMM. FUNCTIONS
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In the first integral in the second part,
we make the substitution u = –x .
Then, du = –dx, and when x = –a, u = a.
INTEGS. OF SYMM. FUNCTIONS0
0
0 0
( ) ( ) ( )
( ) ( )
a a
a a
a a
f x dx f x dx f x dx
f x dx f x dx
Proof
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Therefore,
0 0
0
( ) ( )( )
( )
a a
a
f x dx f u du
f u du
ProofINTEGS. OF SYMM. FUNCTIONS
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So, Equation 8 becomes:
0 0
( )
( ) ( )
a
a
a a
f x dx
f u du f x dx
Proof—Equation 9INTEGS. OF SYMM. FUNCTIONS
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If f is even, then f(–u) = f(u).
So, Equation 9 gives:
0 0
0
( )
( ) ( )
2 ( )
a
a
a a
a
f x dx
f u du f x dx
f x dx
INTEGS. OF SYMM. FUNCTIONS Proof a
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If f is odd, then f(–u) = –f(u).
So, Equation 9 gives:
INTEGS. OF SYMM. FUNCTIONS Proof b
0 0
( )
( ) ( )
0
a
a
a a
f x dx
f u du f x dx
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Theorem 7 is
illustrated here.
INTEGS. OF SYMM. FUNCTIONS
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For the case where
f is positive and
even, part (a) says
that the area under
y = f(x) from –a to a
is twice the area
from 0 to a because
of symmetry.
INTEGS. OF SYMM. FUNCTIONS
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Recall that an integral can be
expressed as the area above the x–axis
and below y = f(x) minus the area below
the axis and above the curve.
INTEGS. OF SYMM. FUNCTIONS
( )b
af x dx
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Therefore, part (b) says the integral
is 0 because the areas cancel.
INTEGS. OF SYMM. FUNCTIONS
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As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even.
So,
2 26 6
2 0
2717 0
1287
2847
( 1) 2 ( 1)
2
2 2
x dx x dx
x x
Example 10INTEGS. OF SYMM. FUNCTIONS
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As f(x) = (tan x)/ (1 + x2 + x4) satisfies
f(–x) = –f(x), it is odd.
So, 1
2 41
tan0
1
xdx
x x
Example 11INTEGS. OF SYMM. FUNCTIONS