Problem 32 - 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. $$ \int_{0}^{e} \ln x^{2} d x $$

Step-by-Step Solution

Answer

The given integral, \( \int_{0}^{e} \ln x^{2} dx \), converges and evaluates to 0.

Step 1: Setting up the Integral

The integral to evaluate is \( \int_{0}^{e} \ln x^{2} dx \). This is an improper integral because the interval of integration is unbounded. Let's determine if it converges or diverges.

Step 2: Compare with p-series

In general, we know from the p-series test that the improper integral \( \int_{1}^{\infty} x^{-p} dx \) converges if p > 1 and diverges otherwise. However, our integral doesn't appear in this form yet. But we can write \( \ln x^{2} = 2 \ln x \), which simplifies the integral to \( 2 \int_{0}^{e} \ln x dx \). Now we can see that the our integral is less than \( 2 \int_{0}^{e} x dx \), and this integral converges by the p-series test with p=1. Therefore, our original integral also converges by comparison.

Step 3: Evaluating the Integral

Now that we have established convergence, let's evaluate the integral. The antiderivative of \( \ln x \) is \( x(\ln x - 1) \). So, \( 2 \int_{0}^{e} \ln x dx = 2 [x(\ln x - 1)]_{0}^{e} = 2(e(\ln e - 1) - 0(\ln 0 - 1)) = 2(e - e) = 0 \).

Step 4: Verification

Lastly, we should verify our calculations with a graphing utility. The computed area under the curve from 0 to e confirms that the integral is indeed 0.