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Math Help - improper integral

  1. #1
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    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
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  2. #2
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    \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
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  3. #3
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    should i be getting a negative answer for this..?

    i got -8/3
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  4. #4
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    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.
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  5. #5
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    would you mind telling me how to do that? :x


    edit:

    i got it, nevermind, thank you very much!
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