Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{1} \frac{1}{x^{2}} d x $$

Short answer

The improper integral \( \int_{0}^{1} \frac{1}{x^{2}} dx \) diverges.

Step-by-step solution

Identify the type of the improper integral

The integral \(\int_{0}^{1} \frac{1}{x^{2}} d x\) is an improper integral because the function \( \frac{1}{x^{2}} \) is not defined at \(x=0\). This kind of issue creates a singularity, which needs special attention.

Set up limit to evaluate integral

Because of the singularity at \(x=0\), express the given improper integral as a limit: \( \int_{0}^{1} \frac{1}{x^{2}}\, dx = \lim_{{t\to 0^+}}\int_{{t}}^{1} \frac{1}{x^{2}}\, dx \) . Here, the notation \( t \to 0^+ \) represents the right-hand limit, indicating that \( t \) approaches zero from positive values.

Evaluate the limit and integral separately

First, compute the integral as: \[ \int_{t}^{1} x^{-2}\, dx = [-x^{-1}]_{t}^{1} = -\frac{1}{1} + \frac{1}{t} \] Now, compute the limit: \[ \lim_{{t \to 0^+}} \left(-1 + \frac{1}{t} \right) \]

Determine convergence / divergence using limit

According to the limit computation rules, having \( t \) in the denominator approaching zero results in positive infinity, i.e., \( \lim_{ {t \to 0^+}} \frac{1}{t} = +\infty \) Thus, the integral converges to infinity, which implies that the original integral is in fact divergent.