476
CHAPTER 7 TECHNIQUES OF INTEGRATION
In the notation for indenite integrals this equation becomes
y
t x f x dx
yf
or
f xt x
x t x dx
ytxf
f xtx
x dx
f xtx
We can rearrange this equation as
yf
1
x t x dx
f xtx
ytxf
x dx
Formula 1 is called the formula f
408
CHAPTER 5 INTEGRALS
Applications
Part 2 of the Fundamental Theorem says that if f is continuous on a, b , then
y
b
f x dx
a
Fb
Fa
where F is any antiderivative of f. This means that F
ten as
y
b
a
F x dx
Fb
f , so the equation can be rewritFa
We know
404
CHAPTER 5 INTEGRALS
60.
y
y
y
f
y=
0.4
y=
0.2
0
1
3
5
7
t
9
_0.2
s
61 62
s
s
s
s
s
s
s
s
s
s
n
nl
62. lim
nl
s
i1
i3
n4
1
n
s
1
n
s
2
n
s
3
n
s
s
n
n
s
s
s
s
s
s
1
t
0.
64. If f is continuous and t and h are differentiable functions, nd
a formula for
400
CHAPTER 5 INTEGRALS
So the equation of FTC2 can be written as
y
b
a
f x dx
Other common notations are F x
Fx
b
a
]
b
where
a
b
a
and F x
x 2 from 0 to 1.
1
3
x 2 is F x
using Part 2 of the Fundamental Theorem:
SOLUTION An antiderivative of f x
y
A
1
0
402
CHAPTER 5 INTEGRALS
 5.3
Exercises
1. Explain exactly what is meant by the statement that differenti
ation and integration are inverse processes.
x0x f
t dt, where f is the function whose graph
is shown.
(a) Evaluate t x for x 0, 1, 2, 3, 4, 5, and
396
CHAPTER 5 INTEGRALS
Intuitively, we therefore expect that
tx
tx
lim
h
h
hl0
tx
fx
The fact that this is true, even when f is not necessarily positive, is the rst part of the
Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus, Part 1
394
CHAPTER 5 INTEGRALS
4. Suppose f is a continuous function on the interval a, b and we dene a new function t by
the equation
y
tx
x
a
f t dt
Based on your results in Problems 1 3, conjecture an expression for t x .
 5.3
The Fundamental Theorem of Calc
398
CHAPTER 5 INTEGRALS
Figure 7 shows the graphs of f x
sin x 2 2 and the Fresnel function
x
Sx
x0 f t dt. A computer was used to graph S by computing the value of this integral
for many values of x. It does indeed look as if S x is the area under the gr
392
CAS
CHAPTER 5 INTEGRALS
3132  Express the integral as a limit of sums. Then evaluate,
using a computer algebra system to nd both the sum and the limit.
31.
y
s
32.
sin 5x dx
0
s
s
s
s
s
y
10
2
s
evaluate x13 2e x
s
s
s
s
(c)
2
0
y
7
5
f x dx
(b)
y
f
390
CHAPTER 5 INTEGRALS
val a, b . In this case Property 8 says that the area under the graph of f is greater than the
area of the rectangle with height m and less than the area of the rectangle with height M .
Proof of Property 8 Since m
M , Property 7 g
382
CHAPTER 5 INTEGRALS
y
y=
+
+
0a
b
_
x
If f takes on both positive and negative values, as in Figure 3, then the Riemann sum is
the sum of the areas of the rectangles that lie above the xaxis and the negatives of the areas
of the rectangles that lie b
388
CHAPTER 5 INTEGRALS
Properties of the Integral
1.
2.
area=c(ba)
0
a
b
b
y
f+g
g
b
a
b
a
b
a
c dx
cb
fx
a , where c is any constant
t x dx
y
b
a
y
f x dx
b
t x dx
a
b
c y f x dx, where c is any constant
c f x dx
a
fx
t x dx
y
b
a
y
f x dx
b
t x dx
a
P
378
CHAPTER 5 INTEGRALS
Because Equation 5 has the same form as our expressions for area in Equations 2 and
3, it follows that the distance traveled is equal to the area under the graph of the velocity
function. In Chapters 6 and 8 we will see that other
384
CHAPTER 5 INTEGRALS
and the right endpoints are x 1 0.5, x 2
x 6 3.0. So the Riemann sum is
1.0, x 3
1.5, x 4
2.0, x 5
2.5, and
6
R6
y
f xi
f 0.5
5
x
i1
y=6x
1
2
x
2.875
f 1.0
f 1.5
x
5.625
5
x
f 2.0
4
0.625
x
f 2.5
x
f 3.0
x
9
3.9375
0
x
3
Notice th
386
CHAPTER 5 INTEGRALS
EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas.
(a)
y
1
0
s1
x 2 dx
(b)
y
3
0
x
1 dx
SOLUTION
y
(a) Since f x
s1 x 2 0, we can interpret this integral as the area under the curve
y s1 x 2 from 0 t
380
CAS
CHAPTER 5 INTEGRALS
x 5 from 0 to 2 as a
23. (a) Express the area under the curve y
CAS
limit.
(b) Use a computer algebra system to nd the sum in your
expression from part (a).
(c) Evaluate the limit in part (a).
CAS
25. Find the exact area under
372
CHAPTER 5 INTEGRALS
Thus
2
1
n
1
n
1
n
Rn
1
n
12
1
n2
12
1
n3
2
2
n
22
22
1
n
2
3
n
32
1
n
n
n
2
n2
32
n2
Here we need the formula for the sum of the squares of the rst n positive integers:
The ideas in Examples 1 and 2 are
explored in Module 5.1/5.2/
376
CHAPTER 5 INTEGRALS
y
1
and x* 1.75, and the sum of the areas of the four approximating rectangles (see Fig4
ure 14) is
y=e
4
f x*
i
M4
x
i1
0
1
2
f 0.25
x
e
FIGURE 14
1
2
0.25
e
x
f 0.75
0.5
0.25
0.75
e
e
0.75
x
0.5
e
f 1.25
e
1.25
1.25
x
0.5
e
1.7
406
CHAPTER 5 INTEGRALS
Table of Indefinite Integrals
1
y cf
c y f x dx
x dx
y k dx
kx
n
dx
xn 1
n1
ye
x
dx
ex
2
1
2
1
tan x
dx
n
y
1
1
dx
x
ya
C
y sec x tan x dx
yx
C
cos x
y sec x dx
fx
yf
t x dx
x dx
ytx
dx
C
yx
y sin x dx
y
ax
ln a
dx
C
C
y cos x dx
C
410
CHAPTER 5 INTEGRALS
(b) Note that v t
t2 t 6
t 3 t 2 and so v t
0 on the interval 1, 3
and v t
0 on 3, 4 . Thus, from Equation 3, the distance traveled is
y
 To integrate the absolute value of v t , we
use Property 5 of integrals from Section 5.2 to
416
CHAPTER 5 INTEGRALS
possible, try choosing u to be some complicated part of the integrand. Finding the right
substitution is a bit of an art. Its not unusual to guess wrong; if your rst guess doesnt
work, try another substitution.
EXAMPLE 2 Evaluate
S
The volume of a sphere is the
limit of sums of volumes of
approximating cylinders.
A pplications of Integration
In this chapter we explore some of the applications of the
denite integral by using it to compute areas between
curves, volumes of solids, and
496
CHAPTER 7 TECHNIQUES OF INTEGRATION
 7.4
Integration of Rational Functions by Partial Fractions
In this section we show how to integrate any rational function (a ratio of polynomials) by
expressing it as a sum of simpler fractions, called partial fra
588
CHAPTER 9 DIFFERENTIAL EQUATIONS
P
Lets try to think of a solution of Equation 1. This equation asks us to nd a function
whose derivative is a constant multiple of itself. We know that exponential functions have
that property. In fact, if we let P t
C
552
CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
Thus, the arc length function is given by
x
ft
1
y
x
2
1
8t
y s1
sx
dt
2t
1
1
8
x2
ln x
dt
1
8
t2
x
]
ln t
1
1
For instance, the arc length along the curve from 1, 1 to 3, f 3 is
1
8
32
s3
ln 3
1
ln 3
8
8
440
CHAPTER 6 APPLICATIONS OF INTEGRATION
(mi/h)
EXAMPLE 4 Figure 8 shows velocity curves for two cars, A and B, that start side by side
and move along the same road. What does the area between the curves represent? Use
the Midpoint Rule to estimate it.
518
CHAPTER 7 TECHNIQUES OF INTEGRATION
 7.7
Approximate Integration
There are two situations in which it is impossible to nd the exact value of a denite
integral.
The rst situation arises from the fact that in order to evaluate xab f x d x using the
Fun
576
CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
Every continuous random variable X has a probability density function f . This means
that the probability that X lies between a and b is found by integrating f from a to b :
Pa
1
X
y
b
b
a
f x dx
For examp
426
CHAPTER 5 INTEGRALS
e x is an increas. Thus, e
0 for all x.
Properties of the Exponential Function The exponential function f x
ing continuous function with domain and range 0,
Also
lim e x 0
lim e x
xl
x
xl
e x.
So the xaxis is a horizontal asymptot
430

CHAPTER 5 INTEGRALS
5 Review
s
CONCEPT CHECK
1. (a) Write an expression for a Riemann sum of a function f .
Explain the meaning of the notation that you use.
(b) If f x
0, what is the geometric interpretation of a
Riemann sum? Illustrate with a dia