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Thread: convergent and divergent improper integrals

  1. #1
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    convergent and divergent improper integrals

    Did I do this one right? :

    $\displaystyle \int \frac{4}{\sqrt{6-x}}dx$ [0,6]

    $\displaystyle = lim \int \frac{4}{\sqrt{6-x}}dx$ [0,6]
    $\displaystyle x->6^-$

    $\displaystyle = lim (\frac{4}{(6-x)^2})| [0,b]$
    $\displaystyle x->6^-$

    $\displaystyle = infinity - \frac{1}{9} = diverges?$
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  2. #2
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    Quote Originally Posted by saiyanmx89 View Post
    Did I do this one right? :

    $\displaystyle \int \frac{4}{\sqrt{6-x}}dx$ [0,6]

    $\displaystyle = lim \int \frac{4}{\sqrt{6-x}}dx$ [0,6]
    $\displaystyle x->6^-$

    $\displaystyle = lim (\frac{4}{(6-x)^2})| [0,b]$
    $\displaystyle x->6^-$

    $\displaystyle = infinity - \frac{1}{9} = diverges?$
    your antiderivative is incorrect ...

    $\displaystyle \int \frac{4}{\sqrt{6-x}} \, dx = -8\sqrt{6-x} + C$
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