Unit 6Integrals with Infinite Limits or Unbounded Integrands
Concept 4.74Built for May 2027 AP Exam

Integrals with Infinite Limits or Unbounded Integrands

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

An improper integral has either an infinite limit or an integrand that blows up inside the interval, and it is defined as a limit of ordinary integrals. Nothing is evaluated at infinity; instead the finite endpoint is replaced by a variable, the ordinary integral is computed, and then the limit is taken. The answer is a number when the limit exists and the integral is said to converge; otherwise the integral diverges. The benchmark family is 1xp\frac{1}{x^{p}} over [1,∞)[1,\infty), which converges exactly when p>1p > 1, and knowing that one comparison settles most multiple-choice items. This topic sits in CED 6.13 and is marked BC only, so it appears on the BC form rather than the AB form.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see
On the BC form, an item that asks you to evaluate an improper integral or show that it diverges, usually with an infinite upper limit and an integrand that came from an earlier part of the same question. On the AB form, nothing: this topic is not tested.
What you’ll do
Replace the offending limit with a variable, integrate normally, then take a limit. When the integrand is unbounded at an interior point, split the integral at that point and require both pieces to converge.
Memory hook
Never plug in infinity. Take a limit instead. The integral sign with an infinite bound is shorthand for a limit written elsewhere.
Use it on the exam
For MCQs: Compare to 1xp\frac{1}{x^{p}}. If the integrand decays faster than 1x\frac{1}{x} at infinity, the integral converges; slower, and it diverges. The comparison needs no antiderivative. For FRQs: Write the limit notation explicitly, as lim⁡b→∞∫abf\lim_{b\to\infty}\int_a^b f. The limit is the definition, and a paper that evaluates at ∞\infty directly has skipped the scored step even when the number is right.

Concept Review

Two Ways an Integral Goes Bad

Figure 1 — The region runs on forever to the right, yet its area can still be finite. The integral is defined as the limit of the shaded area as the right edge moves out.

An integral is improper in either of two situations. The interval is unbounded, as in ∫a∞f\int_a^{\infty} f, or the integrand is unbounded somewhere on the interval, as in ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx where the integrand blows up at the left endpoint. Both cases are handled the same way: cut the integral at a finite point, compute the ordinary integral, then take a limit.

∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\,dx = \lim_{b\to\infty}\int_a^{b} f(x)\,dx

SituationDefinitionWhat can go wrong
infinite upper limitlim⁡b→∞∫abf\lim_{b\to\infty}\int_a^b fthe limit fails to exist
infinite lower limitlim⁡a→−∞∫abf\lim_{a\to-\infty}\int_a^b fsame
both limits infinitesplit at any finite point, both must convergeone piece diverging is enough
unbounded at an endpointlim⁡c→a+∫cbf\lim_{c\to a^{+}}\int_c^b fthe one-sided limit fails
unbounded insidesplit at the bad pointboth pieces must converge
The region runs on forever to the right, yet its area can still be finite. The integral is defined as the limit of the shaded area as the right edge moves out.

The Benchmark Family

∫1∞1xp dx converges when p>1, and diverges when p≤1\int_1^{\infty}\frac{1}{x^{p}}\,dx \ \text{converges when } p > 1, \text{ and diverges when } p \le 1

The two cases worth memorizing by hand are p=2p = 2, which converges to 11, and p=1p = 1, which diverges. The computation shows why the boundary sits at 11:

∫1b1x2 dx=1−1b→1,∫1b1x dx=ln⁡b→∞\int_1^{b}\frac{1}{x^{2}}\,dx = 1 - \frac{1}{b} \to 1, \qquad \int_1^{b}\frac{1}{x}\,dx = \ln b \to \infty

In the first case the antiderivative approaches a finite value; in the second it grows without bound. Every comparison item reduces to deciding which side of p=1p = 1 the integrand sits on.

An Unbounded Integrand That Still Converges

∫011x dx=lim⁡c→0+[2x]c1=2−0=2\int_0^{1}\frac{1}{\sqrt{x}}\,dx = \lim_{c\to 0^{+}}\left[2\sqrt{x}\right]_c^{1} = 2 - 0 = 2

The region is unbounded in height and has finite area. That is the counterintuitive result the topic is built to teach: an infinite spike can enclose a finite amount, just as an infinite tail can. Divergence is a property of the decay rate, not of the picture looking large.

Reading Convergence Off the Integrand

Integrand near the trouble pointBehaviorWhy
1x2\frac{1}{x^{2}} as x→∞x\to\inftyconvergesdecays faster than 1x\frac{1}{x}
1x\frac{1}{x} as x→∞x\to\inftydivergesthe boundary case
1x\frac{1}{\sqrt{x}} as x→∞x\to\inftydivergesdecays slower than 1x\frac{1}{x}
1x\frac{1}{\sqrt{x}} as x→0+x\to 0^{+}convergesthe exponent is less than 11 at a finite endpoint
1x\frac{1}{x} as x→0+x\to 0^{+}divergeslogarithmic growth at the endpoint

The last two rows are the mirror image of the first two, and mixing them up is the standard error. At infinity a larger exponent helps; at a finite endpoint a larger exponent hurts.

Why the Limit Form Matters

Writing ∫a∞\int_a^{\infty} and treating ∞\infty as a number produces answers that are right by accident and wrong as reasoning. The limit form also decides partial credit: a paper that writes lim⁡b→∞\lim_{b\to\infty} has shown the definition, and a paper that writes [−1x]a∞\left[-\frac{1}{x}\right]_a^{\infty} has written something that does not parse. On the BC form, where this topic is scored, the notation is part of the answer.

Common Misconception Check

  • Myth 1: “An infinite region has infinite area.” Truth: The tail can thin out fast enough to leave a finite area, and ∫1∞x−2=1\int_1^{\infty} x^{-2} = 1 is the standard example.
  • Myth 2: “A vertical asymptote always makes the integral diverge.” Truth: ∫01x−1/2=2\int_0^1 x^{-1/2} = 2. The exponent decides, not the asymptote.
  • Myth 3: “If one piece of a split integral diverges, the total is still definable.” Truth: One divergent piece makes the whole integral divergent. There is no cancellation across an infinite quantity.

Topic-Specific Study Method

Replace, Integrate, Limit

Borrowed idea: Three moves, always in the same order. Replace the infinite bound with a variable. Integrate the ordinary integral that results, and take the limit last. The order is what keeps the notation honest, and the notation is scored. Skipping the third move is the commonest error, and it happens because the first two moves feel like the whole computation. A paper that never writes the limit has skipped the definition, even when the number is right.

Exam disguiseWhat you see on the pageWhat your brain must translateRequired execution
∫1∞1x2\int_1^{\infty}\frac{1}{x^{2}}an infinite upper bounda limit in bbreplace, integrate, limit
“Determine whether it converges”no value requesteda comparison is enoughcheck the exponent against 11
Denominator vanishes insidea root in the intervala split at that rootboth pieces must converge
An earlier part produced g(x)g(x)a multi-part stemthe integral is the last stepreuse the antiderivative already found
“Show that it diverges”a proof requestedexhibit a limit that failsshow the antiderivative grows without bound

① Write the limit before anything else. The notation is not decoration; it is the definition, and on the BC form it is scored.

② Compare the exponent to 11 for infinite limits and to 11 again for endpoint blow-ups, remembering that the direction of the test flips between the two cases.

③ When the integrand is unbounded inside the interval, split at the bad point first. A single integral across an asymptote is not defined, no matter what the antiderivative looks like.

Training & Error Diagnosis

Core Archetype (Simplest Form): Evaluate ∫1∞1x3 dx\int_1^{\infty}\frac{1}{x^{3}}\,dx.

Standard Execution:

  1. Replace. lim⁡b→∞∫1bx−3 dx\lim_{b\to\infty}\int_1^{b} x^{-3}\,dx.
  2. Integrate. The antiderivative is −12x2-\frac{1}{2x^{2}}.
  3. Evaluate. [−12x2]1b=−12b2+12\left[-\frac{1}{2x^{2}}\right]_1^{b} = -\frac{1}{2b^{2}} + \frac{1}{2}.
  4. Limit. As b→∞b\to\infty the first term vanishes, leaving 12\frac{1}{2}.
  5. Conclude. The integral converges to 12\frac{1}{2}.

Exam Variation (Context / Parameter Shift): Evaluate ∫0414−x dx\int_0^{4}\frac{1}{\sqrt{4-x}}\,dx, or show that it diverges.

Solution: The integrand blows up at x=4x = 4, so the integral is lim⁡c→4−∫0c(4−x)−1/2dx\lim_{c\to 4^{-}}\int_0^{c}\left(4-x\right)^{-1/2}dx. An antiderivative is −24−x-2\sqrt{4-x}, so the definite integral is −24−c+24=4−24−c-2\sqrt{4-c} + 2\sqrt{4} = 4 - 2\sqrt{4-c}. As c→4−c\to 4^{-} the radical vanishes and the value is 44. The integral converges to 44. The step that carries the credit is the one-sided limit, since the approach is from below and writing c→4c\to 4 without a direction hides the reason the endpoint is excluded.

Common Traps & Failure Signals:

  1. Infinity substituted (Signal: an expression with ∞\infty treated as a number): use a limit.
  2. Exponent test reversed (Signal: a claim that ∫01x−2\int_0^1 x^{-2} converges): at a finite endpoint, a larger exponent makes things worse.
  3. Split omitted (Signal: one integral written across an interior asymptote): the integral does not exist until it is split.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

  • Limit Skipped: Write the limit line first.
  • Picture Trusted: You decided divergence from the shape of the region, which fails in both directions, because a tall narrow spike can enclose finite area while a shallow wide tail can enclose infinite area, and only the exponent separates the two cases.
  • Exponent Test Misdirected: You applied the infinity rule at a finite endpoint, so a convergent integral was declared divergent, and the error is invisible because the reasoning sounds identical in both settings.

Past 10-Year AP Exam Appearance

Testing Frequency
Rare in FRQs
This topic is CED 6.13 and marked BC only, so it never appears on the AB form; on the BC form one part in the released sets since 2017 asks for an improper integral
10-Year Total Count
1 of 9 BC Exams · 1 FRQ Part
On the AB form the count is zero, because improper integrals are outside the AB course description
FRQ Years Appeared
2023 (Q5b) ---

Practice Examples

MCQ 1

The Boundary Case

Determine the convergence of the improper integral ∫1∞1x dx\int_1^{\infty}\frac{1}{x}\,dx.

MCQ 2

An Unbounded Integrand

Determine whether the improper integral ∫011x2 dx\int_0^{1}\frac{1}{x^{2}}\,dx converges.

MCQ 3

A Convergent Spike

Evaluate the improper integral ∫011x dx\int_0^{1}\frac{1}{\sqrt{x}}\,dx and choose the correct value.

FRQ 4

Two Improper Integrals and a Decision

(a) Evaluate ∫1∞1x4 dx\int_1^{\infty}\frac{1}{x^{4}}\,dx, showing the limit that defines it.

(b) Evaluate ∫091x dx\int_0^{9}\frac{1}{\sqrt{x}}\,dx, or show that it diverges.

(c) A student claims that ∫1∞1x0.9 dx\int_1^{\infty}\frac{1}{x^{0.9}}\,dx converges because the integrand approaches zero. Determine whether the claim is correct, and justify your answer.

Cheatsheet

Core Reference

DefinitionTest
∫a∞f\int_a^{\infty} flim⁡b→∞∫abf\lim_{b\to\infty}\int_a^b flimit must exist
∫−∞bf\int_{-\infty}^{b} flim⁡a→−∞∫abf\lim_{a\to-\infty}\int_a^b fsame
unbounded at aalim⁡c→a+∫cbf\lim_{c\to a^{+}}\int_c^b fone-sided limit
unbounded insidesplit at the pointboth pieces converge
∫1∞x−p\int_1^{\infty} x^{-p}benchmarkconverges iff p>1p > 1
∫01x−p\int_0^{1} x^{-p}benchmarkconverges iff p<1p < 1

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic PracticeNext Topic: 6.10 The Trapezoidal Rule formula for approximating the area under a curve.