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Calculus 1 Problems & Solutions – Chapter 6 – Section 6.7.1 |
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6.7.1
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Review |
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1. Proper And Improper Integrals |
Let the function f be continuous on the closed finite
interval [a,
b]
([a,
b]
is finite 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. 1.1f 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
infinite 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
finite 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 Infinite Intervals |
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Fig. 2.1Here,
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Fig. 2.2 |
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Fig. 2.3 |
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 |
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 finite 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 |
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
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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. 5.5Improper Integral For Example 5.1. |
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. 8.1Integral 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. 9.1This unbounded region has an infinite area. |
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Fig. 9.2This unbounded region has a finite area. |
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Fig. 9.3This 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 x-axis or along the y-axis or both, and whether it
extends to infinity on 1 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
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:
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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