Calculus Of One Real Variable –
By Pheng Kim Ving 
11.2 
1. Proper And Improper Integrals 
Let the function f be
continuous on the closed bounded interval [a, b] ([a, b] is bounded if both a
and b are (finite) numbers).
See Fig. 1.1. Under these conditions, f attains
both a maximum and a minimum values (see Section
1.2.2 Theorem 2.1),
Fig. 1.1
f is continuous on [a, b],
is a (finite) number.

In this section we're going to extend our study of definite integrals
to include those of functions that are continuous on
unbounded intervals and of functions that are discontinuous at a finite number
of points. Such definite integrals will be called
improper integrals. Definite integrals of functions continuous on closed
bounded intervals are called proper integrals.
When we say that f is
continuous on [a, b),
we mean that f is: (1) continuous on (a, b), (2)
rightcontinuous at a, and
(3) not leftcontinuous (and thus is discontinuous) at b.
See Figs. 3.1, 3.2,
and 3.3. Similarly for when we say that f is
continuous on (a, b]
or (a, b).
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2. Improper Integrals Over HalfOpen Unbounded Intervals 
Fig. 2.1
Here, 
Fig. 2.2

Fig. 2.3

This observation motivates the definitions of convergent
and divergent improper integrals, as seen in the
following
definition.

Determine if each of the following improper integrals converges. Sketch a graph in each case.
a.
b.
EOS
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Fig. 3.1

Fig. 3.2

Fig. 3.3

Fig. 3.4


Determine whether each of the following improper integrals converges. Sketch a graph in each case.
a.
b.
EOS
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4. Improper Points And The 4 Basic Types Of Improper Integrals 
Each of the 4 improper integrals defined in Definitions 2.1 and 3.1 has
only 1 improper point. There's only 1 limit to
handle for each of them. So each of them is said to be of a basic type.
An improper integral is said to be of a basic type
if it has only 1 improper point. As we'll see later on in this section, all
other types are based on these basic ones. Note
that an improper integral is a finite number if it exists.
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^{{5.1}} Part 3.
Now let f be continuous
on the open bounded interval (a, b). Refer to Fig. 5.4. The definite integral of f over (a, b) is also an
improper integral; it has 2 improper points, a and
b,
not just 1. Again there are 3 situations for the behavior of f at or
near each of a and b.
So there are 6 situations. We show the situation where f
is undefined at a and b
and unbounded
near them in Fig. 5.4. The improper integral of f
we're going to define now is the same for all 6 situations. Let c be an
arbitrary point such that a < c < b. The
improper integral of f over (a, b) is defined
to be the sum of the basictype
improper integrals of f over the
halfopen finite intervals (a, c] and [c, b), and converges iff both the basic types
converge.
Fig. 5.1

Fig. 5.2

Fig. 5.3

Fig. 5.4

In each case, the improper integral on the lefthand side
converges iff both the basictype improper integrals on the 
Remark that the point c should be
chosen so that it makes the calculations of the improper integrals on the
righthand
side as simple as possible.
Determine whether or not the integral:
converges. See Fig. 5.5.
Fig. 5.5
Improper Integral For Example 5.1.

Solution
EOS
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6. NonBasic Types Of Improper Integrals 
Each of the improper integrals defined in Definition 5.1 has 2 improper points. Each is of
a nonbasic type. We say that an
improper integral is of a nonbasic type if it has more than 1 improper
points. There are more than 1 limits to handle
for a nonbasictype improper integral.
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7. Breaking NonBasicType Improper Integrals 
Part 5 shows the necessity
that nonbasictype improper integrals must be broken into (ie, expressed as a
sum of )
separate
basictype improper integrals, and the way to break them. There we break the
given improper integrals into 2 basic types.
A nonbasictype improper integral will be broken into basic
types. There are nonbasic types that must be broken into
more than 2 basic types.
Determine whether the following improper integral converges or diverges. Sketch a graph.
Solution
EOS
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8. Examining Definite Integrals For Improper Points 
Find:
See Fig. 8.1.
Fig. 8.1
Integral For Example 8.1. 
Correct Solution
The integrand 1/x is undefined & so is discontinuous at x = 0. Thus:
EOS
Recall that the fundamental theorem of calculus applies only
if the interval of integration is finite and closed of the form
[a, b]
and the integrand is continuous on [a, b]. In the incorrect solution, the theorem is
incorrectly applied to a function
that is not continuous at 0 and hence not continuous on [–1, 1].
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9. Areas Of Unbounded Regions 
The region under the graph of y
= 1/x, above the xaxis,
and to the right of the vertical line x = 1,
colored in Fig. 9.1,
extends to infinity on the righthand side. It's an unbounded region. Its area is
found in Example 2.1.a and is infinite. This
unbounded region has an infinite area.
The region under the graph of y
= 1/x^{2}, above the xaxis,
and to the right of the vertical line x = 1,
colored in Fig. 9.2,
extends to infinity on the righthand side. It's an unbounded region. Its area
is found in Example 2.1.b and is 1. This
unbounded region has a finite area.
^{{9.1}} Example 5.1.
Fig. 9.1
This unbounded region has an infinite
area.

Fig. 9.2
This unbounded region has a finite area.

Fig. 9.3
This unbounded region has a finite area.

In general, some unbounded regions have infinite areas while
others have finite areas. This is true whether a region
extends to infinity along the xaxis or
along the yaxis or both, and whether it
extends to infinity on one or both sides of an
axis.
Problems & Solutions

1. Determine whether each of the following improper integrals converges, and find its value if it does.
Solution
d. Using the method of integration by parts let u = ln x and dv = dx, so that du = dx/x and v = x. Thus:
convergent to –1.
2.
Solution
a.
3. Evaluate each of the following improper integrals or show that it diverges.
Solution
d. Utilizing the method of integration by parts let u = x and dv = e^{–}^{x} dx, so that du = dx and v = – e^{–}^{x}. Consequently:
e. We have:
4.
a. Sketch the curve y =
e^{x}.
Shade the region that lies above the xaxis, below
the curve y = e^{x}, and to the left of the
yaxis.
b. Find the area of the shaded
region.
Solution
a.
5. Prove that, for a > 0, the improper integral:
Solution
If p = 1 then we have:
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