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How do you evaluate an integral with an infinite limit or an unbounded integrand using limits?

Topic 6.13 Evaluating Improper Integrals: evaluate integrals with infinite limits of integration or an infinite discontinuity by rewriting them as limits of proper integrals, determining convergence or divergence (BC).

A focused answer to AP Calculus BC Topic 6.13, evaluating improper integrals with infinite limits or unbounded integrands by replacing the bad endpoint with a limit, and deciding convergence versus divergence, with worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The two types and the limit definition
  3. A worked type-1 integral
  4. A worked type-2 integral
  5. The p-integral benchmark
  6. Splitting when both ends are improper

What this topic is asking

The College Board (Topic 6.13, BC only) extends the definite integral to two cases the Fundamental Theorem does not directly cover: an infinite limit of integration (the interval runs to ±∞\pm\infty) and an infinite discontinuity in the integrand (the function blows up somewhere on the interval). Both are handled the same way, by replacing the troublesome endpoint with a variable and taking a limit.

The two types and the limit definition

A worked type-1 integral

A worked type-2 integral

The p-integral benchmark

A family worth memorizing is ∫1∞1xp dx\int_1^{\infty} \frac{1}{x^p}\,dx, which converges if p>1p > 1 and diverges if p≤1p \le 1. The borderline case p=1p = 1 gives ∫1∞1x dx=lim⁡b→∞ln⁡b=∞\int_1^\infty \frac{1}{x}\,dx = \lim_{b\to\infty}\ln b = \infty, which diverges, while p=2p = 2 converges to 11. The mirror-image fact near zero is ∫011xp dx\int_0^{1} \frac{1}{x^p}\,dx, which converges if p<1p < 1 and diverges if p≥1p \ge 1. These two benchmarks let you predict the behavior of many improper integrals at a glance and connect directly to the p-series test in Unit 10, where ∑1np\sum \frac{1}{n^p} obeys the same p>1p > 1 rule.

Splitting when both ends are improper

If an integral is improper at both ends, or has a discontinuity in the interior, you must split it at a convenient point so each piece has exactly one source of trouble, then require every piece to converge. For ∫−∞∞f(x) dx\int_{-\infty}^{\infty} f(x)\,dx, write it as ∫−∞0f+∫0∞f\int_{-\infty}^{0} f + \int_0^{\infty} f and take two separate limits; the original converges only if both do, and its value is the sum. The frequent error is integrating across an interior infinite discontinuity as if it were proper, ignoring the blow-up, which gives a meaningless finite number. Always scan the integrand for points where it is undefined inside the interval, not just at the stated endpoints.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (BC, style)1 marksSection I (multiple choice, no calculator). ∫1∞1x2 dx=\int_1^{\infty} \frac{1}{x^2}\,dx = (A) diverges (B) 11 (C) 12\frac{1}{2} (D) 00
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The correct answer is (B), 11.

∫1∞x−2 dx=lim⁡b→∞∫1bx−2 dx=lim⁡b→∞[−1x]1b=lim⁡b→∞(−1b+1)=1\int_1^{\infty} x^{-2}\,dx = \lim_{b\to\infty}\int_1^b x^{-2}\,dx = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b = \lim_{b\to\infty}\left(-\frac{1}{b} + 1\right) = 1. The limit exists, so the integral converges to 11.

AP 2023 (BC, style)4 marksSection II (free response, no calculator). (a) Determine whether ∫011x dx\int_0^{1} \frac{1}{\sqrt{x}}\,dx converges, and if so find its value. (b) Determine whether ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\,dx converges.
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A 4-point convergence problem with one of each improper type.

(a) (2 points) The integrand is unbounded at x=0x = 0. ∫01x−1/2 dx=lim⁡a→0+∫a1x−1/2 dx=lim⁡a→0+[2x]a1=lim⁡a→0+(2−2a)=2\int_0^1 x^{-1/2}\,dx = \lim_{a\to 0^+}\int_a^1 x^{-1/2}\,dx = \lim_{a\to 0^+}\left[2\sqrt{x}\right]_a^1 = \lim_{a\to 0^+}(2 - 2\sqrt{a}) = 2. It converges to 22.
(b) (2 points) ∫1∞1x dx=lim⁡b→∞[ln⁡x]1b=lim⁡b→∞ln⁡b=∞\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b\to\infty}[\ln x]_1^b = \lim_{b\to\infty}\ln b = \infty. It diverges.

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