Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{5}^{\infty} \frac{x}{\sqrt{x^{2}-16}} d x $$

Short answer

The integral diverges.

Step-by-step solution

Evaluate the Integral

To evaluate this integral, we can first perform a substitution where \(u = x^{2}-16\), \(du = 2x \, dx\), and \(dx = \frac{du}{2x}\). So, we get \\[\begin{aligned}&=\int_{5}^{\infty} \frac{x}{\sqrt{u}} \frac{du}{2x} \&=\frac{1}{2}\int_{5}^{\infty} \frac{1}{\sqrt{u}} du.\end{aligned}\]

Solve the New Integral

Now we can solve the integral, by thinking of it as integral of \(u^{-1/2}\), which can be easily integrated as \[\begin{aligned}&=\frac{1}{2} \cdot 2u^{1/2} \&=\sqrt{u}\end{aligned}\]We will evaluate this from 5 to infinity.

Substitute \(u\) back in terms of \(x\)

Substitute \(u\) back into the integral as \(u = x^{2}-16\). So, our expression becomes: \[\begin{aligned}&=\sqrt{x^{2}-16}. \\end{aligned}\]

Evaluate limit as \(x\) goes to infinity

The final step is to evaluate the expression from 5 to infinity.\[\begin{aligned}&= \lim_{x \to \infty} \sqrt{x^{2}-16} - \sqrt{5^{2}-16} \&= \lim_{x \to \infty} \sqrt{x^{2}-16} - \sqrt{9} \&= \lim_{x \to \infty} \sqrt{x^{2}-16} - 3 \&= \infty - 3 \&= \infty\\end{aligned}\]So, the integral diverges.