improper integral

**buttonbear** · March 8th 2009, 11:13 AM

okay, so the problem is:

integral from 2 to 6 y/(y-2)^1/2 dy

i know you have to start by saying lim a-->2+ integral from a to 6 y/(y-2)^1/2

but then i'm kind of blanking on how to integrate/go from there

**skeeter** · March 8th 2009, 11:38 AM

$\int \frac{y}{\sqrt{y-2}} \, dy$

let $u = y - 2$

$du = dy$

$y = u + 2$

$\int \frac{u+2}{\sqrt{u}} \, du$

$\int u^{\frac{1}{2}} + 2u^{-\frac{1}{2}} \, du$

integrate, back substitute, then evaluate your improper integral

**buttonbear** · March 8th 2009, 06:32 PM

should i be getting a negative answer for this..?

i got -8/3

**GaloisTheory1** · March 8th 2009, 07:25 PM

it is possible, you can get a negative answer

and since it is a proper integral, you can check the answer even with a ti-83 plus.

**buttonbear** · March 8th 2009, 07:29 PM

would you mind telling me how to do that? :x

edit:

i got it, nevermind, thank you very much!

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