How do I know if an integral is improper?

Check two things. First, does any limit of integration equal ∞ or -∞? That makes it Type 1. Second, does the integrand become undefined or unbounded anywhere on the interval, usually where a denominator equals zero? That makes it Type 2. If either is true, you must use a limit.

What is the difference between convergence and divergence?

After rewriting the integral as a limit and evaluating, if you get a finite number the integral converges to that number. If the limit is ±∞ or does not exist, the integral diverges. Convergence means the region has finite total area even if it extends infinitely.

Can I cancel infinities in an integral like _-∞^∞ x\,dx?

No. You must split it into two independent improper integrals and require each to converge on its own. In _-∞^∞ x\,dx, each half diverges, so the whole diverges — even though the pieces look like they should cancel by symmetry.

Do I really need to write the limit notation on the AP exam?

Yes. Free-response rubrics typically award a specific point for correctly expressing the improper integral as a limit before evaluating. Jumping straight to plugging in ∞ can cost you credit even if your final number is right.

AP-CALCBC-6.13

U6.13 Improper Integrals

Master AP Calculus BC improper integrals: infinite limits and discontinuous integrands. Learn to rewrite with limits and decide convergence vs. divergence.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.13 Improper Integrals, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

An improper integral looks like an ordinary definite integral, but something breaks the usual rules: either a limit of integration runs off to infinity, or the integrand blows up somewhere on the interval. You cannot just plug numbers into an antiderivative, because you would be evaluating at infinity or at a point where the function is undefined.

The fix is elegant and testable: replace the trouble spot with a variable, integrate normally, then take a limit. If that limit is a finite number, the integral converges; if it is infinite or does not exist, it diverges. This lesson shows you exactly how to set up the limit, evaluate it cleanly, and justify convergence in the language AP graders expect.

Two Types of Improper Integrals

An integral is improper for one of two reasons, and you must diagnose which before doing anything else.

Type 1 (infinite interval): at least one limit of integration is

\infty

-\infty

. Examples:

\int_1^\infty \frac{1}{x^2}\,dx

\int_{-\infty}^0 e^{x}\,dx

.

Type 2 (discontinuous integrand): the function has an infinite discontinuity (vertical asymptote) somewhere on the interval, possibly at an endpoint or interior point. Example:

\int_0^1 \frac{1}{\sqrt{x}}\,dx

is improper because the integrand blows up at

x=0

.

A single integral can be improper for both reasons, like

\int_0^\infty \frac{1}{\sqrt{x}(1+x)}\,dx

, which is improper at

0

and at

\infty

Feature	Type 1	Type 2
Cause	limit is $\pm\infty$	integrand undefined/unbounded
Clue	$\infty$ in a bound	denominator $=0$ on interval
Fix	let bound $\to \pm\infty$	let variable $\to$ bad point

The most common exam mistake is not noticing Type 2. If you see

\int_{-1}^{1}\frac{1}{x^2}\,dx

and blindly apply the Fundamental Theorem, you get a negative answer for a positive function — a red flag that a hidden discontinuity at

x=0

was ignored.

The Limit Setup That Earns the Points

Every improper integral must be rewritten as a limit before you touch an antiderivative. This is not optional style — AP scoring rubrics award a point specifically for the correct limit expression.

For an infinite upper limit:

\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^b f(x)\,dx.

For a discontinuity at the upper endpoint

b

\int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx.

When the problem is improper at both ends, or has an interior discontinuity, split it at a convenient point

c

into two separate improper integrals:

\int_{-\infty}^{\infty} f\,dx = \int_{-\infty}^{c} f\,dx + \int_{c}^{\infty} f\,dx.

The crucial rule: the whole integral converges only if every piece converges. If any piece diverges, the entire integral diverges — you cannot let opposite infinities cancel. Similarly for an interior asymptote at

x=c

inside

[a,d]

, split into

\int_a^c

and

\int_c^d

, each handled with its own one-sided limit. Never integrate straight across a vertical asymptote.

Evaluating the Limit and Deciding Convergence

After setting up the limit, find the antiderivative

F(x)

using techniques from earlier in the unit (substitution, parts, partial fractions), evaluate at the variable bound, and then take the limit.

Key limit behaviors you will use constantly: as

b\to\infty

e^{-b}\to 0

\frac{1}{b^p}\to 0

for

p>0

, and

\ln b\to\infty

. If the resulting limit is a finite real number, write "the integral converges to that value." If the limit is

\pm\infty

or fails to exist (for example

\lim_{b\to\infty}\cos b

), write "the integral diverges."

A hugely useful benchmark is the p-integral:

\int_1^\infty \frac{1}{x^p}\,dx \text{ converges if } p>1,\ \text{diverges if } p\le 1.

Near zero the behavior flips:

\int_0^1 \frac{1}{x^p}\,dx

converges if

p<1

and diverges if

p\ge 1

. Memorizing these lets you predict answers and check your work instantly. The borderline case

p=1

always diverges, because its antiderivative is

\ln x

, which grows without bound at

\infty

and dives to

-\infty

0

How the Exam Tests This Topic

On multiple choice, you will often just evaluate a converging integral or identify which of several integrals diverges — the p-integral rule saves time here. On free response, presentation matters: show the limit notation explicitly, evaluate the antiderivative, and state convergence with the final value.

A classic trap uses functions like

\frac{1}{x^2}

over an interval containing zero, testing whether you notice the discontinuity. Another common setup pairs improper integrals with integration by parts, such as

\int_0^\infty x e^{-x}\,dx

, where you must handle a

\lim_{b\to\infty} b e^{-b}

term (which is

0

by growth rates).

Comparison arguments occasionally appear: if

0\le f(x)\le g(x)

and

\int g

converges, then

\int f

converges. While full comparison tests belong more to series, the intuition helps you predict outcomes.

Common misconceptions to avoid: assuming a bounded-looking region always has finite area (it may not), forgetting that divergence of one split piece dooms the whole integral, and dropping the limit notation and losing setup credit. Always double-check the sign and magnitude of your answer against the geometry — a positive integrand cannot yield a negative area.

Key terms

Improper Integral.: A definite integral with an infinite limit of integration or an integrand that becomes unbounded on the interval of integration.
Converge.: An improper integral converges if its defining limit exists and equals a finite real number.
Diverge.: An improper integral diverges if its defining limit is infinite or does not exist.
Type 1 Improper Integral.: An improper integral caused by an infinite limit of integration, such as $\int_a^\infty f\,dx$ .
Type 2 Improper Integral.: An improper integral caused by an infinite discontinuity in the integrand at an endpoint or interior point.
p-integral.: The reference integral $\int_1^\infty x^{-p}\,dx$ , which converges when $p>1$ and diverges when $p\le 1$ .
Splitting Rule.: For integrals improper at two spots, break into pieces; the whole converges only if every piece converges.

Worked example

Determine whether

\int_0^\infty x e^{-x}\,dx

converges, and if so, find its value.

This is Type 1 because the upper limit is infinite. Rewrite with a limit:

\int_0^\infty x e^{-x}\,dx = \lim_{b\to\infty}\int_0^b x e^{-x}\,dx.

Use integration by parts with

u=x

dv=e^{-x}\,dx

, so

du=dx

and

v=-e^{-x}

\int x e^{-x}\,dx = -x e^{-x} - \int -e^{-x}\,dx = -x e^{-x} - e^{-x} + C.

Evaluate from

0

b

\left[-x e^{-x} - e^{-x}\right]_0^b = \left(-b e^{-b} - e^{-b}\right) - \left(0 - 1\right) = -b e^{-b} - e^{-b} + 1.

Now take the limit as

b\to\infty

. Since exponential decay beats polynomial growth,

b e^{-b}\to 0

and

e^{-b}\to 0

. Therefore

\lim_{b\to\infty}\left(-b e^{-b} - e^{-b} + 1\right) = 0 - 0 + 1 = 1.

The limit is finite, so the integral converges to

1

Practice questions

Which of the following integrals diverges?

$\int_1^\infty \frac{1}{x^2}\,dx$
$\int_1^\infty \frac{1}{x^{3/2}}\,dx$
$\int_1^\infty \frac{1}{x}\,dx$
$\int_0^1 \frac{1}{\sqrt{x}}\,dx$

Answer: $\int_1^\infty \frac{1}{x}\,dx$

By the p-integral rule on

[1,\infty)

\int_1^\infty x^{-p}\,dx

converges only when

p>1

. Here

p=1

, so it diverges (its antiderivative

\ln x\to\infty

). The choices with

p=2

and

p=3/2

converge. The last choice,

\int_0^1 x^{-1/2}\,dx

, is improper at

0

with

p=1/2<1

, so it converges to

2

Evaluate

\int_0^1 \frac{1}{\sqrt{x}}\,dx

or explain why it diverges. Show the limit setup.

Answer: The integral converges to $2$ .

The integrand is unbounded at

x=0

, so this is a Type 2 improper integral. Write

\lim_{t\to 0^+}\int_t^1 x^{-1/2}\,dx

. The antiderivative is

2\sqrt{x}

, so the definite integral equals

2\sqrt{1}-2\sqrt{t}=2-2\sqrt{t}

. As

t\to 0^+

2\sqrt{t}\to 0

, giving a limit of

2

. Because the limit is finite, the integral converges to

2

. This matches the near-zero p-rule since

p=1/2<1

Explain why

\int_{-1}^{1}\frac{1}{x^2}\,dx

cannot be evaluated by simply computing

\left[-\frac{1}{x}\right]_{-1}^{1}

, and determine whether it converges.

Answer: It diverges.

The integrand

\frac{1}{x^2}

has an infinite discontinuity at

x=0

, which lies inside

[-1,1]

, so the Fundamental Theorem cannot be applied straight across. Naively computing gives

-1-1=-2

, which is impossible for a positive integrand. Split at

0

\int_0^1 x^{-2}\,dx=\lim_{t\to0^+}\left[-\frac{1}{x}\right]_t^1=\lim_{t\to0^+}\left(-1+\frac{1}{t}\right)=\infty

. Since this piece diverges, the entire integral diverges.

FAQ

How do I know if an integral is improper?: Check two things. First, does any limit of integration equal $\infty$ or $-\infty$ ? That makes it Type 1. Second, does the integrand become undefined or unbounded anywhere on the interval, usually where a denominator equals zero? That makes it Type 2. If either is true, you must use a limit.
What is the difference between convergence and divergence?: After rewriting the integral as a limit and evaluating, if you get a finite number the integral converges to that number. If the limit is $\pm\infty$ or does not exist, the integral diverges. Convergence means the region has finite total area even if it extends infinitely.
Can I cancel infinities in an integral like $\int_{-\infty}^{\infty} x\,dx$ ?: No. You must split it into two independent improper integrals and require each to converge on its own. In $\int_{-\infty}^{\infty} x\,dx$ , each half diverges, so the whole diverges — even though the pieces look like they should cancel by symmetry.
Do I really need to write the limit notation on the AP exam?: Yes. Free-response rubrics typically award a specific point for correctly expressing the improper integral as a limit before evaluating. Jumping straight to plugging in $\infty$ can cost you credit even if your final number is right.

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