Test convergence of an improper integral

Primary Tool

Math Workspace (CAS)

CAS is the right fit because the example moves directly from antiderivatives to the limiting step that decides convergence. The transcript keeps the convergent and divergent comparison side by side.

Problem

Test whether $\int_{1}^{\infty} \frac{1}{x ^{2}} d x$ converges, and if it does, find its value.

This is a standard improper-integral example because the antiderivative is simple, but the important step is interpreting the infinite interval as a limit.

Why the limit decides everything

An improper integral is not evaluated by writing down an antiderivative and stopping. You first replace the infinite endpoint with a variable cutoff $b$ , evaluate on $[1, b]$ , and then send $b$ to infinity.

That is why $\int_{1}^{\infty} \frac{1}{x ^{2}} d x$ converges while $\int_{1}^{\infty} \frac{1}{x} d x$ diverges, even though both have antiderivatives you can write down immediately.

Step-by-step walkthrough

The computation is short, but every line matters because the classification as convergent or divergent happens at the limit stage.

1Rewrite the integral as $lim_{b \to \infty} \int_{1}^{b} x^{- 2} d x$ .
2Find an antiderivative: $\int x^{- 2} d x = - 1/ x + C$ .
3Evaluate the finite-interval integral to get $\int_{1}^{b} x^{- 2} d x = 1 - 1/ b$ .
4Take the limit as $b \to \infty$ and obtain $lim_{b \to \infty} (1 - 1/ b) = 1$ .
5Conclude that the improper integral converges and its value is $1$ .
6Compare with $\int_{1}^{\infty} \frac{1}{x} d x$ : the antiderivative is $lo g x$ , and $lo g b \to \infty$ , so that improper integral diverges.

Common Pitfall

Finding an antiderivative is not enough. Improper integrals are classified only after you evaluate the truncated integral and take the limit at the problematic endpoint.

Try a Variation

Test $\int_{1}^{\infty} \frac{1}{x ^{p}} d x$ for $p = 3$ and for $p = 1/2$ . Which step changes, and what pattern do you notice?

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