# Question 2e22f

Refer to explanation

#### Explanation:

I would start by making a substitution u=10−x, and du=−dx# . Then it becomes

${\int}_{\infty}^{0} \frac{1}{\sqrt{u}} \left(- 1\right) \mathrm{du}$
which you can rewrite as

${\int}_{0}^{\infty} \frac{1}{\sqrt{u}} \mathrm{du}$

This is an improper integral for two reasons. First, it's got an infinite limit of integration. Second, it's infinite at $u = 0$. It's probably going to be infinite for one of those two reasons. But let's go ahead and see what we get. The antiderivative of $\frac{1}{\sqrt{u}}$ is $2 \sqrt{u}$. So we get

${\left[2 \sqrt{u}\right]}_{0}^{\infty} = 2 \sqrt{\infty} - 2 \sqrt{0} \to \infty$

So the final answer is that the integral diverges to $\infty$.