AP Calculus BC

Definite vs Improper Integral

Q: Can a region of infinite extent have finite area?

Yes, and that is the point of convergence. The region under y = \frac{1}{x^2} from 1 to infinity never ends, yet its integral is exactly 1.

Q: How does this connect to series?

Through the integral test. For a continuous, positive, decreasing f with f(n) = a_n, the series \sum_{n=1}^{\infty} a_n and the integral \int_1^{\infty} f(x)\,dx converge or diverge together, which is where the p-series rule comes from.

An integral is improper when a limit of integration is infinite or the integrand blows up somewhere on the interval. You replace the bad endpoint with a variable, evaluate an ordinary definite integral, and take a limit. If that limit is finite the integral converges; otherwise it diverges and has no value.

Definite

Use when: Both bounds are finite and the integrand stays bounded on the interval, so the value is an ordinary number and no limit is needed.

Improper

Use when: A bound is infinite, or the integrand has a vertical asymptote at an endpoint or inside the interval. Improper integrals are CED topic 6.13, which is BC only.

Side by side

	Definite	Improper
What it measures	Exact accumulation over a finite interval	A limit of accumulations over growing or shrinking intervals
Trigger	Finite bounds and a bounded integrand	An infinite bound, or an infinite discontinuity on $[a,b]$
Written as	$\int_a^b f(x)\,dx$	$\lim_{t \to \infty} \int_a^t f(x)\,dx$
Possible outcomes	Always a number	Converges to a number, or diverges
Common trap	Not noticing the integrand is unbounded near an endpoint	Evaluating straight through an asymptote instead of splitting

Improper is a statement about the setup, not about difficulty. Check two things before integrating anything: does a bound read $\infty$ or $-\infty$ , and does the integrand have a vertical asymptote anywhere on the interval, endpoints included. Either one makes the integral improper and obliges you to write the limit.

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

The $p$ integral is the benchmark worth memorizing. $\int_1^{\infty} \frac{1}{x^p}\,dx$ converges exactly when $p > 1$ , so $\int_1^{\infty} \frac{1}{x^2}\,dx = 1$ while $\int_1^{\infty} \frac{1}{x}\,dx$ diverges. Near zero the inequality flips, because there the danger is the asymptote rather than the tail: $\int_0^1 \frac{1}{x^p}\,dx$ converges exactly when $p < 1$ .

Split at an interior asymptote

When the integrand blows up strictly inside the interval, break the integral there and take a one-sided limit on each side; both pieces must converge for the whole to converge. Pushing $\int_{-1}^{1} \frac{1}{x^2}\,dx$ through the Fundamental Theorem returns $-2$ , which is impossible for a positive integrand. The integral in fact diverges.

Frequently asked questions

Can a region of infinite extent have finite area?

Yes, and that is the point of convergence. The region under $y = \frac{1}{x^2}$ from $1$ to infinity never ends, yet its integral is exactly $1$ .

How does this connect to series?

Through the integral test. For a continuous, positive, decreasing $f$ with $f(n) = a_n$ , the series $\sum_{n=1}^{\infty} a_n$ and the integral $\int_1^{\infty} f(x)\,dx$ converge or diverge together, which is where the $p$ -series rule comes from.

In the CED: Unit 6: Integration and Accumulation