AP Calculus AB: 12.3.2 The Second Type of Improper Integral
This content explains the second type of improper integral, which occurs when the integrand is discontinuous within the interval of integration. To evaluate such integrals correctly, the discontinuity must be treated as a boundary point. The interval is split at the discontinuity, and limits are used to evaluate each subinterval. Proper setup is essential; failing to recognize the discontinuity can lead to incorrect results, especially if the function changes behavior or is undefined at certain points.
The Second Type of Improper Integral
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.
Key Terms
The Second Type of Improper Integral
If a function is not continuous on the integration interval, then the standard procedure will not work. Use the discontinuity as an endpoint for th...
note
Remember:
A definite integral is considered an improper integral if it has one of these properties:
· the integration is ...
Evaluate ∫1 0 x^−1/4 dx
None of the above
What is the correct way to evaluate the improper integral
∫ 4 −4 1/x^4 dx
using subintervals?
∫0−41x4dx +∫401x4dx
Consider ∫2−21x3dx.
For which values of x is the integrand discontinuous?
0
Evaluate ∫ 5 1 1/(x−4)^2dx.
The integral diverges.
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| Term | Definition |
|---|---|
The Second Type of Improper Integral | 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. |
note |
|
Evaluate ∫1 0 x^−1/4 dx | None of the above |
What is the correct way to evaluate the improper integral using subintervals? | ∫0−41x4dx +∫401x4dx |
Consider ∫2−21x3dx. For which values of x is the integrand discontinuous? | 0 |
Evaluate ∫ 5 1 1/(x−4)^2dx. | The integral diverges. |
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? | Most of the area is below the x‑axis. |
Evaluate ∫1−11x2dx. | The integral diverges. |
Evaluate ∫π02sec2xdx. | The integral diverges. |
What is the correct way to evaluate the improper integral using subintegrals? | ∫1.00.5−1x(lnx)2dx+∫1.51.0−1x(lnx)2dx |