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Math Help - Improper Integrals

  1. #1
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    Unhappy 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.
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  2. #2
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    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
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  3. #3
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    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?
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  4. #4
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    Re: Improper Integrals

    Quote Originally Posted by kashmoneyrecord3 View Post
    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 ?
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  5. #5
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    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
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  6. #6
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    Re: Improper Integrals

    Quote Originally Posted by kashmoneyrecord3 View Post
    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.
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  7. #7
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    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?
    Last edited by kashmoneyrecord3; October 13th 2011 at 05:33 PM.
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  8. #8
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    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.
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