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

(Doh)

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- Mar 8th 2009, 11:13 AMbuttonbearimproper integral
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

(Doh) - Mar 8th 2009, 11:38 AMskeeter
$\displaystyle \int \frac{y}{\sqrt{y-2}} \, dy$

let $\displaystyle u = y - 2$

$\displaystyle du = dy$

$\displaystyle y = u + 2$

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

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

integrate, back substitute, then evaluate your improper integral - Mar 8th 2009, 06:32 PMbuttonbear
should i be getting a negative answer for this..?

i got -8/3

(Worried) - Mar 8th 2009, 07:25 PMGaloisTheory1
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. - Mar 8th 2009, 07:29 PMbuttonbear
would you mind telling me how to do that? :x

edit:

i got it, nevermind, thank you very much!