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Calculus Of One Real Variable –
By Pheng Kim Ving |
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11.2 |
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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),
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Fig. 1.1
f is continuous on [a, b],
is a (finite) number.
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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)
right-continuous at a, and
(3) not left-continuous (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 Half-Open Unbounded Intervals |
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Fig. 2.1
Here, |
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Fig. 2.2
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Fig. 2.3
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Convergence And Divergence
This observation motivates the definitions of convergent
and divergent improper integrals, as seen in the
following
definition.
Definition 2.1
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Example 2.1
Determine if each of the following improper integrals converges. Sketch a graph in each case.
Solution
a.
b.
EOS
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Fig. 3.1
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Fig. 3.2
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Fig. 3.3
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Fig. 3.4
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Definition 3.1
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Example 3.1
Determine whether each of the following improper integrals converges. Sketch a graph in each case.
Solution
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 basic-type
improper integrals of f over the
half-open finite intervals (a, c] and [c, b), and converges iff both the basic types
converge.
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Fig. 5.1
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Fig. 5.2
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Fig. 5.3
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Fig. 5.4
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Definition 5.1
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In each case, the improper integral on the left-hand side
converges iff both the basic-type improper integrals on the |
Remark that the point c should be
chosen so that it makes the calculations of the improper integrals on the
right-hand
side as simple as possible.
Example 5.1
Determine whether or not the integral:
converges. See Fig. 5.5.
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Fig. 5.5
Improper Integral For Example 5.1.
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Solution
EOS
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6. Non-Basic Types Of Improper Integrals |
Each of the improper integrals defined in Definition 5.1 has 2 improper points. Each is of
a non-basic type. We say that an
improper integral is of a non-basic type if it has more than 1 improper
points. There are more than 1 limits to handle
for a non-basic-type improper integral.
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7. Breaking Non-Basic-Type Improper Integrals |
Part 5 shows the necessity
that non-basic-type improper integrals must be broken into (ie, expressed as a
sum of )
separate
basic-type improper integrals, and the way to break them. There we break the
given improper integrals into 2 basic types.
A non-basic-type improper integral will be broken into basic
types. There are non-basic types that must be broken into
more than 2 basic types.
Example 7.1
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 |
Example 8.1
Find:
See Fig. 8.1.
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Fig. 8.1
Integral For Example 8.1. |
Incorrect Solution
EOS
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 x-axis,
and to the right of the vertical line x = 1,
colored in Fig. 9.1,
extends to infinity on the right-hand 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/x2, above the x-axis,
and to the right of the vertical line x = 1,
colored in Fig. 9.2,
extends to infinity on the right-hand 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.
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Fig. 9.1
This unbounded region has an infinite
area.
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Fig. 9.2
This unbounded region has a finite area.
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Fig. 9.3
This unbounded region has a finite area.
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In general, some unbounded regions have infinite areas while
others have finite areas. This is true whether a region
extends to infinity along the x-axis or
along the y-axis or both, and whether it
extends to infinity on one or both sides of an
axis.
Problems & Solutions
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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 =
ex.
Shade the region that lies above the x-axis, below
the curve y = ex, and to the left of the
y-axis.
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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