Question

Question #2e22f

Answer 1

Refer to explanation

Explanation:

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

`int_(oo)^0 1/sqrtu (-1)du`
which you can rewrite as

`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`. 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 `1/sqrtu` is `2sqrtu`. So we get

`[2sqrtu]_0^oo=2sqrtoo-2sqrt0->oo`

So the final answer is that the integral diverges to `oo`.