# Improper Integrals

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• Oct 13th 2011, 01:03 PM
kashmoneyrecord3
Improper Integrals
Hey guys, I have a question about improper integrals.

For example lets say the question is 1/(x-2)^3/2 dx and the interval is between infinity and 3.

How would I be able to determine if its convergent or divergent and if its convergent solve.

So I tried the problem out myself and what I did was I took the (x-2)^3/2 to the top so it becomes (x-2)^-3/2 and from there i just did the anti derivative of that which i thought was 1/(-3/2+1/1) giving me 1/(-1/4) which i then brought the 4 to the top, meaning the equation is now -4(x-2)^(-3/2+1/1) which turns out to be -4(x-2)^(-1/2) between the interval of infinity and 3. Is what i did so far correct or completely off? If it's wrong can you tell me why I can't do that and what the correct way of doing it is?

Sorry I had to make it long, i just wanted you guys to understand the problem I had.
• Oct 13th 2011, 02:21 PM
skeeter
Re: Improper Integrals
$\int_3^\infty (x-2)^{-\frac{3}{2}} \, dx$

$\lim_{b \to \infty} \int_3^b (x-2)^{-\frac{3}{2}} \, dx$

note the antiderivative is $-2(x-2)^{-\frac{1}{2}}$ ...

$-2 \lim_{b \to \infty} \left[\frac{1}{\sqrt{x-2}}\right]_3^b$

$-2 \lim_{b \to \infty} \left[\frac{1}{\sqrt{b-2}} - 1\right] = 2$

converges
• Oct 13th 2011, 02:34 PM
kashmoneyrecord3
Re: Improper Integrals
oh now I get it, but like for the last line where it says 1/square root of b-2 then -1 right howd u get that to =2? Like how did you manage to get the answer 2? Since its infinity aren't you able to sub in anything meaning lets say 1000. so 1/square root of 1000-2 then -1 doesn't give you 2. So how'd you get that value?
• Oct 13th 2011, 02:47 PM
skeeter
Re: Improper Integrals
Quote:

Originally Posted by kashmoneyrecord3
oh now I get it, but like for the last line where it says 1/square root of b-2 then -1 right howd u get that to =2? Like how did you manage to get the answer 2? Since its infinity aren't you able to sub in anything meaning lets say 1000. so 1/square root of 1000-2 then -1 doesn't give you 2. So how'd you get that value?

I evaluated the limit ... what happens to $\frac{1}{\sqrt{b-2}}$ as $b \to \infty$ ?
• Oct 13th 2011, 02:49 PM
kashmoneyrecord3
Re: Improper Integrals
but like aren't you able to just sub in like say a big number for infinity then evaluate? I'm not sure lol
• Oct 13th 2011, 02:51 PM
skeeter
Re: Improper Integrals
Quote:

Originally Posted by kashmoneyrecord3
but like aren't you able to just sub in like say a big number for infinity then evaluate? I'm not sure lol

... then you need to review how to evaluate limits at infinity.
• Oct 13th 2011, 03:28 PM
kashmoneyrecord3
Re: Improper Integrals
Are you able to quickly summarize it for me? :) I could've swore it was just subbing in a really high number but if it's not how do you solve for it?

Can anyone please tell me how to evaluate using the infinity limit?
• Oct 13th 2011, 05:21 PM
Ackbeet
Re: Improper Integrals
If you're doing improper integrals, you should have already been taught how to evaluate limits at infinity. If you need a review, check your book or another student or your teacher or one of the many fine online tutorials such as this.

Thread closed.