Question #2e22f

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Question #2e22f

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Answer 1 · 2015-09-25T21:41:48

Refer to explanation

Explanation:

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

$\int_{\infty}^{0} \frac{1}{\sqrt{u}} (- 1) d u$ $int_(oo)^0 1/sqrtu (-1)du$
which you can rewrite as

$\int_{0}^{\infty} \frac{1}{\sqrt{u}} d u$ $int_0^oo 1/(sqrtu) du$

This is an improper integral for two reasons. First, it's got an infinite limit of integration. Second, it's infinite at $u = 0$ $u=0$ . 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 $\frac{1}{\sqrt{u}}$ $1/sqrtu$ is $2 \sqrt{u}$ $2sqrtu$ . So we get

${[2 \sqrt{u}]}_{0}^{\infty} = 2 \sqrt{\infty} - 2 \sqrt{0} \to \infty$ $[2sqrtu]_0^oo=2sqrtoo-2sqrt0->oo$

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

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Question #2e22f

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