An improper integral is a definite integral with one or both limits of integration infinite, or with an integrand that is unbounded (has a vertical asymptote) on the interval. You evaluate it by replacing the problem spot with a variable, integrating, then taking a limit (Topic 6.13, BC only).
An improper integral is what you get when a definite integral breaks one of the usual rules. Either the interval runs off to infinity (like ∫₁^∞ 1/x² dx) or the integrand blows up somewhere on the interval (like ∫₀¹ 1/√x dx, where there's a vertical asymptote at x = 0). Per the CED's essential knowledge for 6.13.A, an improper integral "has one or both limits infinite or has an integrand that is unbounded in the interval of integration."
The fix is always the same move. Swap the bad endpoint for a variable, evaluate the definite integral on the bounded piece, then take a limit. So ∫₁^∞ 1/x² dx becomes lim(b→∞) ∫₁^b 1/x² dx, which equals 1. If that limit exists as a finite number, the integral converges to that value. If the limit is infinite or doesn't exist, the integral diverges. The area under the curve goes on forever in one direction, but the total area can still be finite. That's the whole surprising idea.
Improper integrals live in Topic 6.13 (Unit 6: Integration and Accumulation of Change) and are BC only. Learning objective AP Calc 6.13.A asks you to evaluate an improper integral or determine that it diverges, which means showing the limit setup, not just plugging in infinity. The concept comes roaring back in Topic 10.4, where the integral test (AP Calc 10.4.A) uses the convergence of an improper integral to decide whether an infinite series converges or diverges. So this one Unit 6 skill is load-bearing for Unit 10. If you can't handle ∫₁^∞ 1/xᵖ dx, you can't justify why p-series converge for p > 1.
Integral Test for Convergence (Unit 10)
The integral test is the payoff for learning improper integrals. For a positive, decreasing, continuous function, the series Σaₙ and the improper integral ∫₁^∞ f(x) dx converge or diverge together. Every integral test problem is secretly an improper integral problem.
Limits at Infinity and Infinite Limits (Unit 1)
Evaluating an improper integral always ends with a limit, either b→∞ or x approaching a vertical asymptote from one side. Unit 1 limit skills are the final step of every Topic 6.13 problem.
L'Hôpital's Rule (Unit 4)
After you antidifferentiate, the limit you're left with is often an indeterminate form like ∞/∞ (think x·e⁻ˣ as x→∞). L'Hôpital's Rule is the standard tool for finishing those, so the two skills get tested together.
Antiderivatives and Limits of Integration (Unit 6)
An improper integral is still an integral. You need a clean antiderivative before the limit step, and you need to spot when a vertical asymptote sits inside your limits of integration so you can split the integral at that point.
This is a BC-only topic, so AB students can skip it entirely. On MCQs, you'll be asked to evaluate an improper integral, determine that it diverges, or recognize the correct limit setup. Practice questions hit exactly these moves, like identifying that you integrate over a bounded interval and then take a limit, knowing what it means when an integral diverges, and splitting an integral at an interior infinite discontinuity. The classic trap answer treats ∞ like a number you plug in. On FRQs, improper integrals show up inside larger problems. The 2023 BC exam (FRQ 5) asked for one as part of a multi-part question, and the rubric rewards writing the limit notation explicitly, like lim(b→∞) ∫₃^b ... dx. Skipping the limit and just writing ∞ as a bound costs you the point even if your number is right.
Not every discontinuity makes an integral improper. An integrand with a removable discontinuity (a hole) is still bounded, so the definite integral behaves normally and no limit setup is required. An integral is improper only when the integrand is unbounded, meaning there's a vertical asymptote on the interval, or when a limit of integration is infinite. Quick check before you start: does the function blow up, or is it just missing a point? Only the blow-up case is improper.
An integral is improper if one or both limits of integration are infinite, or if the integrand has a vertical asymptote on the interval of integration.
You evaluate an improper integral by replacing the problem endpoint with a variable, integrating, and taking a limit; if the limit is finite the integral converges, and otherwise it diverges.
On FRQs you must write the limit notation explicitly, because plugging ∞ in as a bound without a limit loses the point.
If the vertical asymptote sits inside the interval, split the integral at that point and require both pieces to converge.
The benchmark family ∫₁^∞ 1/xᵖ dx converges when p > 1 and diverges when p ≤ 1, which is the engine behind the p-series rule in Unit 10.
Improper integrals are BC only, appearing in Topic 6.13 and again in Topic 10.4 through the integral test.
It's a definite integral where a limit of integration is infinite or the integrand is unbounded on the interval (CED Topic 6.13). You evaluate it by integrating over a bounded interval and then taking a limit, and the integral either converges to a finite value or diverges.
No. Improper integrals are BC only, covered in Topic 6.13 and used again in Topic 10.4 for the integral test. AB students are not tested on them.
Yes, and this is the big idea of the topic. ∫₁^∞ 1/x² dx converges to exactly 1 even though the region is infinitely long, because the curve shrinks fast enough. Compare that to ∫₁^∞ 1/x dx, which diverges.
A regular definite integral has finite bounds and a bounded integrand, so you can apply the Fundamental Theorem of Calculus directly. An improper integral breaks one of those conditions, so you need an extra limit step. A removable discontinuity does not make an integral improper; only an unbounded integrand or infinite bounds do.
It means the limit of the definite integrals doesn't settle on a finite number, so the integral has no finite value. In Unit 10, divergence of ∫₁^∞ f(x) dx tells you via the integral test that the matching series Σaₙ also diverges.