# improper integral

• Mar 8th 2009, 11:13 AM
buttonbear
improper 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 AM
skeeter
$\displaystyle \int \frac{y}{\sqrt{y-2}} \, dy$

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

i got -8/3
(Worried)
• Mar 8th 2009, 07:25 PM
GaloisTheory1
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 PM
buttonbear
would you mind telling me how to do that? :x

edit:

i got it, nevermind, thank you very much!