LESSON X
LESSON 13 INTEGRALS WITH DISCONTINUOUS INTEGRANDS
Definition If the function f is continuous on the interval [pic] and discontinuous at [pic], then
[pic].
NOTE: Recall that a definite integral exists if the integrand is continuous on the closed interval [pic]of integration. Since the integrand f is continuous on the interval [pic], then it is continuous on the closed interval [pic] for all [pic]. Thus, [pic] exists for all [pic]. Of course, if [pic], then since the function f is continuous at [pic], then the function f is defined at [pic] and [pic].
Definition If the function f is continuous on the interval [pic] and discontinuous at [pic], then
[pic].
Since the integrand f is continuous on the interval [pic], then it is continuous on the closed interval [pic] for all [pic]. Thus, [pic] exists for all [pic]. Of course, if [pic], then since the function f is continuous at [pic], then the function f is defined at [pic] and [pic].
COMMENT: The integral [pic] in the definitions above are also called improper integrals because of the integrand has a discontinuity on the closed interval [pic]. If the limit in the definition exists, we say that the improper integral converges. If the limit does not exist, then we say that the improper integral diverges.
Definition If the function f has a discontinuity at [pic] in the open interval [pic] but is continuous elsewhere in the closed interval [pic], then
[pic] = [pic] + [pic] = [pic] + [pic]
provided that both of the improper integrals converge.
Examples Determine whether the following improper integrals converge or diverge. If the integral converges, then give its value.
1. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the set of real numbers given by [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is the upper limit of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] = 36
Answer: Converges; 36
2. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the interval [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is the lower limit of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] = [pic] = [pic]
Answer: Converges; [pic]
3. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the set of real numbers given by [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at of [pic], which is the upper limit of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] since [pic]
Answer: Diverges
4. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the set of real numbers given by [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is in the interval of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic] + [pic] = [pic] + [pic]
The integrand is continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic] = [pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic] = [pic] since
[pic]
Since the first improper integral [pic] diverges, we do not have to determine whether the second improper integral [pic] converges or diverges. Even if it converges, the improper integral [pic] can not converge.
Answer: Diverges
5. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the set of real numbers given by [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is in the interval of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic] + [pic] =
[pic] + [pic]
The integrand is continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
We will find the partial fraction decomposition for [pic] = [pic]. Thus,
[pic] = [pic].
Multiplying both sides of this equation by [pic], we obtain the following equation.
[pic]
To solve for A, choose [pic]: [pic] [pic]
To solve for B, choose [pic]: [pic] [pic]
Thus, [pic] = [pic] = [pic] =
[pic] = [pic]
Thus,
[pic] = [pic] =
[pic] = [pic] =
[pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] = [pic]
since [pic]
[pic] since [pic]
[pic] since [pic] and [pic] is
finite (It is a finite negative number.)
Thus, the improper [pic] diverges. Thus, the improper [pic] diverges.
Answer: Diverges
6. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the interval [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is the lower limit of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
The integral [pic] is improper with an infinite limit of integration and a discontinuity. However,
[pic] [pic] and
[pic] [pic]
because we want only one limit for the infinite limit of integration and one limit for the discontinuity. In order to accomplish this, we will use the following property of definite integrals.
[pic] = [pic] + [pic],
where c is a real number in the open interval [pic].
Since 3 is in the open interval [pic], then we may write
[pic] = [pic] + [pic] =
[pic] + [pic]
NOTE: You can use any real number in the open interval [pic].
The integrand is continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic]for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] = [pic]
Thus, [pic] converges to [pic].
Now, consider [pic]
[pic] = [pic] = [pic] =
[pic] = [pic] since
[pic]
Thus, [pic] converges to [pic].
Thus, [pic] converges and [pic] =
[pic] + [pic] = [pic] + [pic] =
[pic] = [pic]
Answer: Converges; [pic]
7. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the interval [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has discontinuities at [pic], which is the lower limit of integration, [pic], which is the upper limit of integration, and [pic], which is in the interval. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
NOTE: We need one integral which will handle the discontinuity of [pic] as a lower limit of integration. We need two integrals that will handle the discontinuity of [pic] as an upper limit of integration in one integral and as a lower limit of integration in the other. We need one integral that will handle the discontinuity of [pic] as an upper limit of integration. Thus, we need four integrals. Since [pic] is in the open interval [pic] and 1 is in the open interval [pic], then we may write the following
[pic] = [pic] + [pic] + [pic] +
[pic] = [pic] + [pic] +
[pic] + [pic]
The integrand is continuous on the closed intervals [pic] for all [pic], the closed intervals [pic] for all [pic], the closed intervals [pic] for all [pic], and the closed intervals [pic] for all [pic] . Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
BE CAREFUL: This integral is close to the form of the integral whose answer would involve the inverse secant function since [pic] =
[pic]. We will need to use Trigonometric Substitution in order to evaluate [pic]
Let [pic], where [pic]. Then [pic].
Also, [pic] = [pic] = [pic].
Thus, [pic] since [pic] when [pic]. Thus, we have that
[pic] = [pic] = [pic] = [pic] =
[pic]
Since [pic], then [pic]. Thus, [pic] Using right triangle trigonometry, we have that [pic].
5
t
[pic]
[pic]
Thus, [pic] = [pic] = [pic] =
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] since
[pic]
Thus, the improper integral [pic] converges to [pic].
Now, consider the improper integral [pic].
[pic] = [pic] = [pic] =
[pic] =
[pic]
[pic] = [pic]
[pic] = [pic] = [pic]
Since [pic], then [pic].
Thus, [pic]
Thus, the improper integral [pic] diverges.
Since the improper integral [pic] diverges, then the improper integral [pic] diverges.
Answer: Diverges
NOTE: If you want the practice, you can show that the improper integral [pic] , which implies that it diverges, and the improper integral
[pic] converges to [pic].
8. [pic]
NOTE: The integrand of [pic] has a discontinuity at [pic], which is the lower limit of integration, and is continuous on the interval [pic]. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic]
Since [pic], then [pic].
Answer: Diverges
9. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the set of real numbers given by the open interval [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is the upper limit of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] = [pic] = [pic] = [pic]
since [pic]
Answer: Converges; [pic]
10. [pic]
NOTE: The integrand of [pic] is continuous on its domain of definition, which is the set of real numbers given by the open interval [pic]. The interval of integration is the closed interval [pic]. Thus, the integrand has a discontinuity at [pic], which is the lower limit of integration. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
This integral can be evaluated using Integration by Parts.
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic]
[pic]
[pic] = [pic] = [pic] = [pic]
Thus, [pic] = [pic] = [pic]
Answer: Converges; [pic] or [pic]
11. [pic]
NOTE: The integrand of [pic] has a discontinuity at [pic], which is the lower limit of integration, and is continuous on the interval [pic]. Thus, the Fundamental Theorem of Calculus can not be applied to this integral. We will need to do the following.
[pic] = [pic]
The integrand is continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied to these closed intervals.
[pic]
This integral can be evaluated using Integration by Parts.
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic] =
[pic] =
[pic]
[pic]
[pic] = [pic] = [pic]
Since [pic], then [pic].
Thus, [pic]
Answer: Diverges
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