Improper integrals with infinite discontinuities

About this lesson

Calculus BC (AP-aligned) › Improper integrals

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Improper integrals with infinite discontinuities

Warm-up

Three quick questions from earlier lessons.

Discover

1x\dfrac{1}{\sqrt{x}} blows up at x=0x = 0. Yet ∫t1dxx=2−2t\displaystyle\int_t^1 \dfrac{dx}{\sqrt{x}} = 2 - 2\sqrt{t}. What happens as t→0+t \to 0^+?

  1. It approaches 2
  2. It diverges because 1x\tfrac{1}{\sqrt{x}} is unbounded

Key idea

Approach a vertical asymptote with a limit.

If ff is unbounded at aa: ∫abf(x) dx=lim⁡t→a+∫tbf(x) dx\displaystyle\int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx. If the asymptote is inside [a,b][a, b], split there; the integral converges only if both parts do.

Worked example

∫08dxx3\displaystyle\int_0^8 \dfrac{dx}{\sqrt[3]{x}}.

∫t8x−1/3 dx=6−32t2/3\displaystyle\int_t^8 x^{-1/3}\,dx = 6 - \tfrac{3}{2}t^{2/3}
∫x^(−1/3) dx = (3/2)x^(2/3).
→6\to 6
t → 0⁺.

Your turn, guided

Your turn: ∫04dxx\displaystyle\int_0^4 \dfrac{dx}{\sqrt{x}}.

=4= 4
2√x from t to 4, t → 0.

Practice ★

  1. Evaluate ∫025dxx\displaystyle\int_0^{ 25 } \dfrac{dx}{\sqrt{x} }.
  2. Evaluate ∫01dxx\displaystyle\int_0^{ 1 } \dfrac{dx}{\sqrt{x} }.
  3. Evaluate ∫016dxx\displaystyle\int_0^{ 16 } \dfrac{dx}{\sqrt{x} }.

Watch out!

Practice ★★

  1. What is ∫−22dxx2\displaystyle\int_{-2}^{ 2 } \dfrac{dx}{x^2}?
    1. It diverges: 1x2\dfrac{1}{x^2} is unbounded at x=0x = 0
    2. [−1x]−22=−22\left[-\dfrac{1}{x}\right]_{-2}^{ 2 } = -\dfrac{2}{ 2 }
    3. 0, by symmetry
  1. Evaluate ∫018ln⁡x dx\displaystyle\int_0^1 8\ln x\,dx.
  2. Evaluate ∫015ln⁡x dx\displaystyle\int_0^1 5\ln x\,dx.

Exit ticket

Three questions to check you have got it.

  1. Evaluate ∫025dxx\displaystyle\int_0^{ 25 } \dfrac{dx}{\sqrt{x} }.
  2. Evaluate ∫01dxx\displaystyle\int_0^{ 1 } \dfrac{dx}{\sqrt{x} }.
  3. Evaluate ∫016dxx\displaystyle\int_0^{ 16 } \dfrac{dx}{\sqrt{x} }.

Summary card

Infinite discontinuities

  • Look for asymptotes inside [a,b][a, b] before integrating.
  • Replace the bad end by tt; take the one-sided limit.

Watch out: never apply the FTC across an asymptote.

End of the lesson

Next lesson: Convergence of improper integrals: the p-test Back to the chapter

Answers: Improper integrals with infinite discontinuities

A

  1. 1010
  2. 22
  3. 88

B

  1. It diverges: 1x2\dfrac{1}{x^2} is unbounded at x=0x = 0

C

  1. −8-8
  2. −5-5

D

  1. 1010
  2. 22
  3. 88

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