Problem 29 - Essential Calculus

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 because the limit as \(t\) approaches 0 from the right for \(-\frac{1}{t}\) is \(-\infty\).

Step 1: Identify the Type of Integral

Firstly, identify what kind of integral is given. In this case, we've an improper integral because the function \(f(x) = \frac{1}{x^{2}}\) is undefined at \(x = 0\) which is an endpoint of the interval of integration. We can rewrite it to show where it is undefined: \(\int_{0}^{1} \frac{1}{x^{2}} dx = \lim_{t \to 0^{+}} \int_{t}^{1} \frac{1}{x^{2}} dx = \lim_{t \to 0^{+}}\int_{t}^{1} x^{-2} dx\).

Step 2: Evaluate the Integral

Next, the antiderivative of \(x^{-2}\) needs to be found and then substituted. Antiderivative of \(x^{-2} = -x^{-1}\). Thus, we get: \[ \lim_{t \to 0^{+}} [-x^{-1}]\Big|_{t}^{1} = \lim_{t \to 0^{+}}[-1 - (-\frac{1}{t})]\] Now, evaluate this limit.

Step 3: Evaluate the Limit

The expression within the limit now becomes \(1 - \frac{1}{t}\) as \(t\) approaches 0 from the right. As \(t\) gets smaller and smaller, \(\frac{1}{t}\) gets larger and larger, thus the expression becomes \(-\infty\). Thus, the limit does not exist.