Problem 27 - College Algebra and Calculus: An Applied Approach

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. $$ \int_{0}^{1} \frac{1}{x^{2}} d x $$

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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: \[ \lim_{{t \to 0^+}} \left(-1 + \frac{1}{t} \right) \]

Step 4: 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.