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AP Calculus AB: 12.3.1 The First Type of Improper Integral

Mathematics10 CardsCreated 3 months ago

This content introduces improper integrals, which occur when the interval of integration is infinite or the integrand is undefined within the interval. It explains that some improper integrals can yield a finite area and therefore converge, while others result in an infinite area and diverge. The convergence or divergence depends on the behavior of the function as it approaches the bounds of integration.

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The First Type of Improper Integral

An improper integral is a definite integral with one of the following properties: the integration takes place over an infinite interval, or the integrand is undefined at a point within the interval of integration.
• With some improper integrals, the area of the region under the curve is finite even though the region extends to infinity.
• An improper integral that is infinite diverges. An improper integral that equals a numerical value converges.

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Key Terms

Term
Definition

The First Type of Improper Integral

An improper integral is a definite integral with one of the following properties: the integration takes place over an infinite interval, or the int...

note

  • A definite integral is considered an improper integral if it has one of these properties:
    · the integration is over an infinite interval...

Evaluate ∫∞ 25 250,000/x^3dx.

200

Which of the following best describes the red region under the curve for y = 1 / x^ 2?

The red region is dropping in height as x increases. As a result, the change in the area decreases as x increases

Which of the following statements correctly describes improper integrals?

An improper integral that has a defined numerical value converges. Otherwise the improper integral diverges.

Evaluate ∫∞ 1 1/√x dx

The improper integral diverges.

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AP Calculus AB: 12.3.1 The First Type of Improper Integral
TermDefinition

The First Type of Improper Integral

An improper integral is a definite integral with one of the following properties: the integration takes place over an infinite interval, or the integrand is undefined at a point within the interval of integration.
• With some improper integrals, the area of the region under the curve is finite even though the region extends to infinity.
• An improper integral that is infinite diverges. An improper integral that equals a numerical value converges.

note

  • 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.

  • Imagine calculating the area of the region under the curve y=1/x^2, starting at x = 1 and moving to the right. You would integrate the function from 1 to ∞ and solve the improper integral. The solution of the integral is one. This means that the area under the curve to the right of x = 1 has an area of one. You can think about this area “repackaged” into the square bounded by the origin and the point (1,1). Because this improper integral has a finite value, it converges.

  • An improper integral diverges if its value is infinite. Imagine calculating the area under the curve y=1/x
    , starting at x = 1 and moving to the right. You would integrate the function from 1 to and solve the improper integral. In this example, the value of the integral is infinity, meaning that the area under the curve is infinitely large. This improper integral diverges.

Evaluate ∫∞ 25 250,000/x^3dx.

200

Which of the following best describes the red region under the curve for y = 1 / x^ 2?

The red region is dropping in height as x increases. As a result, the change in the area decreases as x increases

Which of the following statements correctly describes improper integrals?

An improper integral that has a defined numerical value converges. Otherwise the improper integral diverges.

Evaluate ∫∞ 1 1/√x dx

The improper integral diverges.

Which of the following is not an improper integral?

∫−2 0 x dx

Which of the following statements is true for

∫ ∞ 0 e^−x dx?

The integral is improper and converges to 1.

Evaluate ∫ ∞ 0 cosx dx.

The improper integral diverges.

Which of the following best explains why these two improper integrals differ from one another?
∫∞11x2dx∫∞11xdxImproper Integral A Improper Integral B

Improper Integral A is convergent because it approaches a value of 1.
Improper Integral B is divergent because it approaches infinity.