Problem 39 - 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. $$ \int_{0}^{\infty} \frac{4}{\sqrt{x}(x+6)} d x $$

Step-by-Step Solution

Answer

The given improper integral diverges.

Step 1 - Determine if the integral converges or diverges

Compare the integral with \( \int_{0}^{\infty} \frac{1}{x^a} dx \), where \(a<1\). If our original integral is less than this simpler integral, then it converges. Here, we can notice that the integral of \( \int_{0}^{\infty} \frac{4}{\sqrt{x}(x+6)} dx \) will be less than \( \int_{0}^{\infty} \frac{1}{x^a} dx \) for \(a<1\), which implies that our integral converges. We can then move on to compute it.

Step 2 - Compute the integral

We use the substitution method to make the integrand simpler. Letting \( u = x+6 \), we then have \( du = dx \) and \( x = u - 6 \). Our limits, originally from \(0\) to \( \infty \), now change to \(6\) to \( \infty \) with this substitution. The integral becomes \( \int_{6}^{\infty} \frac{4}{\sqrt{u-6}u} du \). This integral can be solved by using a standard integral formula and then applying the limits.

Step 3 - Substitute back the variables and apply limits

Substituting back the variables and simplifying the integral, we get \( \int_{6}^{\infty} [\frac{2}{\sqrt{u}} - \frac{2}{u}] du \) which when integrated and limits are applied gives \( [2\sqrt{u} - 2ln|u|]_{6}^{\infty} = 2\sqrt{\infty} - 2ln|\infty| - 2\sqrt{6} + 2ln|6| \). This simplifies to \( \infty - \infty - 2\sqrt{6} + 2ln|6| \). Now, since infinity minus infinity is indeterminate, we can't proceed further.

Step 4 - Conclude the solution

Thus, while the comparison with a simpler integral suggested that the given integral would converge, upon trying to evaluate it we encountered an indeterminate form. This means that the initial assumption that the integral converged was incorrect and hence the given integral actually diverges.