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Chapter 6: Problem 29

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

Expert verified
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 by step solution

01

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: \(\int_{0}^{1} \frac{1}{x^{2}} dx = \lim_{t \to 0^{+}} \int_{t}^{1} \frac{1}{x^{2}} dx = \lim_{t \to 0^{+}}\int_{t}^{1} x^{-2} dx\).
02

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.
03

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.

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