Question #2e22f

1 Answer
Konstantinos Michailidis
Sep 25, 2015

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#.

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