12.3.2 The Second Type of Improper Integral

Question 1

The Second Type of Improper Integral

Answer 1

• If a function is not continuous on the integration interval, then the standard procedure will not work. Use the discontinuity as an endpoint for the integral. This is the second type of improper integral.

Question 2

note

Answer 2

• Remember:
• A definite integral is considered an improper integral if it has one of these properties:
  · the integration is over an infinite interval; or
  · its integrand is undefined at a point within the
  interval of integration.
• A second type of improper integral occurs when the function is not continuous over the interval of integration. In this example, the function is undefined and the curve is discontinuous at x = 0. This makes the integral improper. In order to get the correct answer, you have to set up the integral so that the point of discontinuity is an endpoint.
• If you didn’t realize that the function is discontinuous at x = 0, you might set up the problem with limits of integration between negative one and one. This would result in a solution with the area under the curve equal to negative two. This answer is incorrect. Notice that the function is squared, so it is positive everywhere; the correct answer must also be positive.
• To set up the integral correctly, use the point of discontinuity, x = 0, as a limit of integration.
• Because the function is symmetric about the y-axis,
  you will need to find the area under the curve between zero and one and then double the result. Evaluating the antiderivative, -2/x , as results in . This improper integral diverges.

Question 3

Evaluate ∫1 0 x^−1/4 dx

Answer 3

None of the above

Question 4

What is the correct way to evaluate the improper integral
∫ 4 −4 1/x^4 dx

using subintervals?

Answer 4

∫0−41x4dx +∫401x4dx

Question 5

Consider ∫2−21x3dx.

For which values of x is the integrand discontinuous?

Answer 5

0

Question 6

Evaluate ∫ 5 1 1/(x−4)^2dx.

Answer 6

The integral diverges.

Question 7

Suppose you have a function y = f (x). You compute the area bound by the function and the x‑axis between −2 and 2 and get a negative answer. What does this negative answer mean?

Answer 7

Most of the area is below the x‑axis.

Question 8

Evaluate ∫1−11x2dx.

Answer 8

The integral diverges.

Question 9

Evaluate ∫π02sec2xdx.

Answer 9

The integral diverges.

Question 10

What is the correct way to evaluate the improper integral
∫1.50.5−1x(lnx)2dx

using subintegrals?

Answer 10

∫1.00.5−1x(lnx)2dx+∫1.51.0−1x(lnx)2dx