5.5 the substitution rule
DESCRIPTION
INTEGRALS. 5.5 The 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. INTRODUCTION. Due to the Fundamental Theorem of Calculus, it is important to be able to find antiderivatives. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/1.jpg)
5.5
The 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
![Page 2: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/2.jpg)
Due to the Fundamental Theorem of Calculus, it is
important to be able to find antiderivatives.
However, our antidifferentiation formulas do not
tell us how to evaluate integrals such as
INTRODUCTION
22 1x x dx Equation 1
![Page 3: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/3.jpg)
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
![Page 4: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/4.jpg)
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. 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
![Page 5: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/5.jpg)
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
![Page 6: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/6.jpg)
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
![Page 7: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/7.jpg)
In general, this method works whenever we have
an integral that we can write in the form
INTRODUCTION
( ( )) ( )f g x g x dx
![Page 8: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/8.jpg)
Observe that, if F’ = f, then
because, by the Chain Rule,
Equation 3INTRODUCTION
( ( )) '( ( )) '( )d
F g x F g x g xdx
( ( )) ( ) ( ( ))F g x g x dx F g x C
![Page 9: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/9.jpg)
That is, 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
![Page 10: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/10.jpg)
Writing F’ = f, we get:
Thus, we have proved the following rule.
INTRODUCTION
( ( )) '( ) ( )f g x g x dx f u du
![Page 11: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/11.jpg)
SUBSTITUTION RULE
If u = g(x) is a differentiable function whose range
is an interval I and f is continuous on I, then
Equation 4
( ( )) '( ) ( )f g x g x dx f u du
![Page 12: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/12.jpg)
SUBSTITUTION RULE
Notice that the Substitution Rule 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.
![Page 13: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/13.jpg)
SUBSTITUTION RULE
Thus, the Substitution Rule says:
It is permissible to operate with dx and du after
integral signs as if they were differentials.
![Page 14: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/14.jpg)
SUBSTITUTION RULE
Find
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
3 4cos( 2)x x dx
![Page 15: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/15.jpg)
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
![Page 16: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/16.jpg)
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
by the simpler integral3 4cos( 2)x x dx1
cos4
u du
![Page 17: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/17.jpg)
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.
![Page 18: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/18.jpg)
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.
![Page 19: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/19.jpg)
Finding the right substitution is a bit of an art.
It is not unusual to guess wrong.
If your first guess does not work, try another substitution.
SUBSTITUTION RULE
![Page 20: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/20.jpg)
SUBSTITUTION RULE
Evaluate
Let u = 2x + 1.
Then, du = 2 dx.
So, dx = du/2.
2 1x dx
E. g. 2—Solution 1
![Page 21: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/21.jpg)
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
![Page 22: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/22.jpg)
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 du udu
![Page 23: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/23.jpg)
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
![Page 24: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/24.jpg)
SUBSTITUTION RULE
Find
Let u = 1 – 4x2. Then, du = -8x dx. So, xdx = -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
![Page 25: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/25.jpg)
SUBSTITUTION RULE
The answer to the example could be checked by
differentiation.
Instead, let us check it with a graph.
![Page 26: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/26.jpg)
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
![Page 27: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/27.jpg)
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
![Page 28: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/28.jpg)
So, it seems reasonable, from the graphical
evidence, that g is an antiderivative of f.
SUBSTITUTION RULE
![Page 29: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/29.jpg)
SUBSTITUTION RULE
Calculate
If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore,
5 15
15
515
x u
u
x
e dx e du
e C
e C
Example 4
5xe dx
![Page 30: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/30.jpg)
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.
5 21x x dx
Example 5
![Page 31: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/31.jpg)
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
![Page 32: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/32.jpg)
SUBSTITUTION RULE
Calculate
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 ln | |
cosln | cos |
x dux dx dx u C
x ux C
Example 6
tan x dx
![Page 33: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/33.jpg)
SUBSTITUTION RULE
Since –ln|cos x| = ln(|cos x|-1)
= ln(1/|cos x|)
= ln|sec x|,
the result can also be written as
Equation 5
tan ln | sec |x dx x C
![Page 34: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/34.jpg)
DEFINITE INTEGRALS
When evaluating a definite integral by substitution,
two methods are possible.
![Page 35: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/35.jpg)
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
![Page 36: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/36.jpg)
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.
![Page 37: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/37.jpg)
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
![Page 38: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/38.jpg)
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
![Page 39: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/39.jpg)
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
![Page 40: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/40.jpg)
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
![Page 41: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/41.jpg)
To find the new limits of integration, we note that: When x = 0, u = 2(0) + 1 = 1 and when x = 4, u = 2(4)
+ 1 = 9
Example 7SUB. RULE FOR DEF. INTEGRALS
![Page 42: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/42.jpg)
Thus,
4 9120 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
![Page 43: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/43.jpg)
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
![Page 44: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/44.jpg)
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
x
Example 8SUB. RULE FOR DEF. INTEGRALS
![Page 45: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/45.jpg)
Thus,2 7
2 21 2
7
2
1
(3 5 ) 5
1 1
5
1 1 1 1
5 7 2 14
dx du
x u
u
Example 8SUB. RULE FOR DEF. INTEGRALS
![Page 46: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/46.jpg)
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
x
Example 9SUB. RULE FOR DEF. INTEGRALS
121
1 00
ln 1
2 2
e x udx u du
x
![Page 47: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/47.jpg)
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
![Page 48: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/48.jpg)
SYMMETRY
The next theorem uses the Substitution Rule for
Definite Integrals to simplify the calculation of
integrals of functions that possess symmetry
properties.
![Page 49: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/49.jpg)
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
![Page 50: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/50.jpg)
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
![Page 51: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/51.jpg)
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. FUNCTIONS
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
![Page 52: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/52.jpg)
Therefore,
0 0
0
( ) ( )( )
( )
a a
a
f x dx f u du
f u du
ProofINTEGS. OF SYMM. FUNCTIONS
![Page 53: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/53.jpg)
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
![Page 54: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/54.jpg)
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
![Page 55: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/55.jpg)
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
![Page 56: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/56.jpg)
Theorem 7 is
illustrated here.
INTEGS. OF SYMM. FUNCTIONS
![Page 57: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/57.jpg)
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
![Page 58: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/58.jpg)
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
![Page 59: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/59.jpg)
Therefore, part (b) says the integral is 0 because
the areas cancel.
INTEGS. OF SYMM. FUNCTIONS
![Page 60: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/60.jpg)
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
![Page 61: 5.5 The Substitution Rule](https://reader036.vdocument.in/reader036/viewer/2022081421/56814830550346895db55155/html5/thumbnails/61.jpg)
As f(x) = (tan x)/ (1 + x2 + x4) satisfies f(–x) = –f(x),
thus f(x) is odd and,
1
2 41
tan0
1
xdx
x x
Example 11INTEGS. OF SYMM. FUNCTIONS