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Thread: improper intergals

  1. #1
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    improper intergals

    OK so i'm having trouble with two problems.

    first off

    $\displaystyle \int_{0}^{2} \frac{1}{\sqrt{l x-1 l}} dx$

    i know to split into two limits but i don't know what to do with the square root.

    second

    $\displaystyle \int_{0}^{1} \frac{1}{2 \sqrt{x}} dx$

    this one i don't really know where to start

    any help would be appreciated, Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Because $\displaystyle |x-1|=|1-x|$ is an 'even function' respect to 1 You need only to compute the first integral between 0 and 1 and then multiply by two. So it is...

    $\displaystyle \int_{0}^{2} \frac{dx}{\sqrt{|x-1|}} = 2\cdot \int_{0}^{1} \frac{dx}{\sqrt{1-x}} = 2\cdot |-2\cdot \sqrt{1-x}|_{0}^{1} = 2\cdot 2 = 4$

    The second integral is similar to the first but easier...

    $\displaystyle \frac{1}{2} \int_{0}^{1}\frac {dx}{\sqrt{x}} = \frac{1}{2} \cdot |2\cdot \sqrt{x}|_{0}^{1} = \frac{1}{2}\cdot 2 = 1$

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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