Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{1}^{\infty} \frac{1}{x^{2}} d x $$

Short answer

The improper integral \( \int_{1}^{\infty} \frac{1}{x^{2}} dx \) converges and its value is 1.

Step-by-step solution

Identify the integral

The integral is an improper integral of the form \( \int_{a}^{\infty} \frac{1}{x^{p}} dx \), with \( p = 2 \) and \( a = 1 \)

Check if the integral converges or diverges

Since \( p = 2 > 1 \), the integral convergence rule tells us that this integral should converge.

Verify the convergence by evaluating the integral

First, rewrite the integral from 1 to infinity as the limit as \( b \to \infty \) of the integral from 1 to \( b \). This gives: \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{2}} dx \). Now evaluate the definite integral: \( \lim_{b \to \infty} [-1/x]_{1}^{b} = \lim_{b \to \infty} (-1/b + 1) \). As \( b \to \infty \), -1/b goes to 0, so the limit becomes \( 0 + 1 = 1 \). Therefore, the integral converges and its value is 1.