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1
1
dx.
x(1 ’ x)1/2
0

SOLUTION
The key idea is that we can only handle one singularity at a time. This
integrand is singular at both endpoints 0 and 1. Therefore we divide the domain
of integration somewhere in the middle”at 1/2 say (it does not really matter
where we divide)”and treat the two singularities separately.
First we treat the integral
1/2 1
dx.
x(1 ’ x)1/2
0
Since the integrand has a singularity at 0, we consider
1/2



Y
1
lim dx.
x(1 ’ x)1/2
’0+
FL
This is a tricky integral to evaluate directly. But notice that
1 1
AM


x(1 ’ x)1/2 x · (1)1/2
¤ x ¤ 1/2. Thus
when 0 <
TE



1/2 1/2 1/2
1 1 1
dx ≥ dx = dx.
x(1 ’ x)1/2 x · (1)1/2 x
We evaluate the integral: it equals ln(1/2) ’ ln . Finally,
lim ’ ln = +∞.
’0+
The ¬rst of our integrals therefore diverges.
But the full integral
1 1
dx
x(1 ’ x)1/2
0
converges if and only if each of the component integrals
1/2 1
dx
x(1 ’ x)1/2
0
and
1 1
dx
x(1 ’ x)1/2
1/2
139
CHAPTER 5 Indeterminate Forms

converges. Since the ¬rst integral diverges, we conclude that the original
integral diverges as well.

’1/3 dx
3
You Try It: Calculate as an improper integral.
’2 (2x)


5.3.3 AN APPLICATION TO AREA
Suppose that f is a non-negative, continuous function on the interval (a, b] which
is unbounded as x ’ a + . Look at Fig. 5.5. Let us consider the area under the graph
of f and above the x-axis over the interval (a, b]. The area of the part of the region
over the interval [a + , b], > 0, is
b
f (x) dx.
a+




Fig. 5.5

Therefore it is natural to consider the area of the entire region, over the interval
(a, b], to be
b
lim f (x) dx.
’0+ a+

This is just the improper integral
b
Area = f (x) dx.
a
140 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.20
Calculate the area above the x-axis and under the curve
1
y= 0 < x ¤ 1/2.
,
x · ln4/3 x

SOLUTION
According to the preceding discussion, this area is equal to the value of the
improper integral
1/2 1/2
1 1
dx = lim dx.
x · ln4/3 x x · ln4/3 x
’0+
0

For clarity we let •(x) = ln x, • (x) = 1/x. Then the (inde¬nite) integral
becomes
3
• (x)
dx = ’ 1/3 .
• 4/3 (x) • (x)

Thus
1/2
1/2 1 3
dx = lim ’
lim
x · ln4/3 x ln1/3 x
’0+ ’0+

’3 ’3
= lim ’ .
[’ ln 2]1/3 [ln ]1/3
’0+

Now as ’ 0 then ln ’ ’∞ hence 1/[ln ]1/3 ’ 0. We conclude that our
improper integral converges and the area under the curve and above the x-axis
equals 3/[ln 2]1/3 .




5.4 More on Improper Integrals
5.4.1 INTRODUCTION
Suppose that we want to calculate the integral of a continuous function f (x) over
an unbounded interval of the form [A, +∞) or (’∞, B]. The theory of the integral
that we learned earlier does not cover this situation, and some new concepts are
needed. We treat improper integrals on in¬nite intervals in this section, and give
some applications at the end.
141
CHAPTER 5 Indeterminate Forms

5.4.2 THE INTEGRAL ON AN INFINITE INTERVAL
Let f be a continuous function whose domain contains an interval of the form
[A, +∞). The value of the improper integral
+∞
f (x) dx
A
is de¬ned to be
N
lim f (x) dx.
N ’+∞ A

Similarly, we have: Let g be a continuous function whose domain contains an
interval of the form (’∞, B]. The value of the improper integral
B
g(x) dx
’∞
is de¬ned to be
B
lim f (x) dx.
M’’∞ M

EXAMPLE 5.21
Calculate the improper integral
+∞
x ’3 dx.
1

SOLUTION
We do this problem by evaluating the limit
N N
x ’3 dx = ’(1/2)x ’2
lim lim
N ’+∞ 1 1
N’+∞

lim ’(1/2) N ’2 ’ 1’2
=
N’+∞
1
=.
2
We conclude that the integral converges and has value 1/2.
EXAMPLE 5.22
Evaluate the improper integral
’32
x ’1/5 dx.
’∞
142 CHAPTER 5 Indeterminate Forms

SOLUTION
We do this problem by evaluating the limit

5 4/5 ’32
’32
’1/5
dx = lim
lim x x
M’’∞ 4
M’’∞ M M
5
= lim (’32)4/5 ’ M 4/5
M’’∞ 4
5
= lim 16 ’ M 4/5 .
M’’∞ 4

This limit equals ’∞. Therefore the integral diverges.
∞ ’3 dx.
1 (1 + x)
You Try It: Evaluate
Sometimes we have occasion to evaluate a doubly in¬nite integral. We do so by
breaking the integral up into two separate improper integrals, each of which can be
evaluated with just one limit.
EXAMPLE 5.23
Evaluate the improper integral
∞ 1
dx.
’∞ 1 + x
2


SOLUTION
The interval of integration is (’∞, +∞). To evaluate this integral, we break
the interval up into two pieces:
(’∞, +∞) = (’∞, 0] ∪ [0, +∞).
(The choice of zero as a place to break the interval is not important; any
other point would do in this example.) Thus we will evaluate separately the
integrals
+∞ 0
1 1
dx and dx.
1 + x2 ’∞ 1 + x
2
0
For the ¬rst one we consider the limit
N 1 N
dx = lim Tan’1 x
lim
1+x 2
N ’+∞ 0 N ’+∞ 0

Tan’1 N ’ Tan’1 0
= lim
N ’+∞
π
= .
2
143
CHAPTER 5 Indeterminate Forms

The second integral is evaluated similarly:
0 1 0
dx = lim Tan’1 x
lim
1+x 2
M’’∞ M M’’∞ M

Tan’1 0 ’ Tan’1 M
= lim
M’’∞
π
= .
2
Since each of the integrals on the half line is convergent, we conclude that the
original improper integral over the entire real line is convergent and that its
value is
π π
+ = π.
2 2
∞ ’1 dx.
1 (1 + x)
You Try It: Discuss

5.4.3 SOME APPLICATIONS
Now we may use improper integrals over in¬nite intervals to calculate area.
EXAMPLE 5.24
Calculate the area under the curve y = 1/[x · (ln x)4 ] and above the x-axis,
2 ¤ x < ∞.
SOLUTION
The area is given by the improper integral
+∞ N
1 1
dx = lim dx.
x · (ln x)4 x · (ln x)4
N ’+∞
2 2
For clarity, we let •(x) = ln x, • (x) = 1/x. Thus the (inde¬nite) integral
becomes
1/3
• (x)
dx = ’ 3 .
• 4 (x) • (x)
Thus
1/3 N
N 1
dx = lim ’ 3
lim
x · (ln x)4
N ’+∞ 2 ln x 2
N’+∞
1/3 1/3
= lim ’ ’3
ln3 N ln 2
N’+∞
1/3
= 3.
ln 2
Thus the area under the curve and above the x-axis is 1/(3 ln3 2).
144 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.25
Because of in¬‚ation, the value of a dollar decreases as time goes on. Indeed,
this decrease in the value of money is directly related to the continuous
compounding of interest. For if one dollar today is invested at 6% contin-
uously compounded interest for ten years then that dollar will have grown
to e0.06·10 = $1.82 (see Section 6.5 for more detail on this matter). This
means that a dollar in the currency of ten years from now corresponds to
only e’0.06·10 = $0.55 in today™s currency.
Now suppose that a trust is established in your name which pays 2t + 50
dollars per year for every year in perpetuity, where t is time measured in
years (here the present corresponds to time t = 0). Assume a constant
interest rate of 6%, and that all interest is re-invested. What is the total
value, in today™s dollars, of all the money that will ever be earned by your
trust account?

SOLUTION
Over a short time increment [tj ’1 , tj ], the value in today™s currency of the
money earned is about

(2tj + 50) · e’0.06·tj · tj .

The corresponding sum over time increments is

(2tj + 50) · e’0.06·tj tj .
j

This in turn is a Riemann sum for the integral

(2t + 50)e’0.06t dt.

If we want to calculate the value in today™s dollars of all the money earned from
now on, in perpetuity, this would be the value of the improper integral
+∞
(2t + 50)e’0.06t dt.
0

This value is easily calculated to be $1388.89, rounded to the nearest cent.

You Try It: A trust is established in your name which pays t + 10 dollars per year
for every year in perpetuity, where t is time measured in years (here the present
corresponds to time t = 0). Assume a constant interest rate of 4%. What is the total
value, in today™s dollars, of all the money that will ever be earned by your trust
account?
145
CHAPTER 5 Indeterminate Forms

Exercises
1. If possible, use l™Hôpital™s Rule to evaluate each of the following limits. In
each case, check carefully that the hypotheses of l™Hôpital™s Rule apply.
cos x ’ 1
(a) lim
x’0 x 2 ’ x 3
e2x ’ 1 ’ 2x
(b) lim
x2 + x4
x’0
cos x
(c) lim
x’0 x 2
[ln x]2
(d) lim
x’1 (x ’ 1)
(x ’ 2)3
(e) lim
x’2 sin(x ’ 2) ’ (x ’ 2)
ex ’ 1
(f ) lim
x’1 x ’ 1
2. If possible, use l™Hôpital™s Rule to evaluate each of the following limits. In
each case, check carefully that the hypotheses of l™Hôpital™s Rule apply.
x3
(a) lim
x’+∞ ex ’ x 2
ln x
(b) lim
x’+∞ x
e’x
(c) lim
x’+∞ ln[x/(x + 1)]

sin x
(d) lim
x’+∞ e’x
ex
(e) lim
x’’∞ 1/x

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