Problem 28 - 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. $$ \int_{0}^{1} \frac{1}{x} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the type of the improper integral

The given integral \(\int_{0}^{1} \frac{1}{x} dx\) is an improper integral, because the integrand \(\frac{1}{x}\) is undefined at \(x = 0\). This results in a vertical asymptote at this point, indicating an infinite discontinuity.

Step 2: Transform the integral into a limit

Since the the integration cannot be done directly due to the singularity at \(x = 0\), let's rewrite the integral as a limit by introducing an auxiliary variable \(a\). Thus, we get \(\lim_{a \to 0^{+}}\int_{a}^{1} \frac{1}{x} dx\). This limit will allow us to investigate what happens as \(x\) approaches 0.

Step 3: Evaluate the limit

The indefinite integral for \(\frac{1}{x}\) is \(\ln|x|\), so the definite integral form \(a\) to \(1\) is \(\ln|1| - \ln|a| = - \ln|a|\). Then, our limit becomes \(\lim_{a \to 0^{+}} - \ln |a|\).

Step 4: Determine if the limit converges or diverges

As \(a\) approaches \(0^{+}\) , the value of \(\ln |a|\) tends towards \(-\infty\). As we have an additional negative sign in front, the limit tends towards \(\infty\). Hence, the improper integral is diverging.