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Math Help - Determine the value of the improper integral

  1. #1
    Junior Member bijosn's Avatar
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    Determine the value of the improper integral

    I = \displaystyle 3 \int_0^{2} \frac{1}{\sqrt{2-x}} dx

    I = \displaystyle -6\sqrt{2-x} \right]_0^{2}

    I = \displaystyle -6\sqrt{0} + 6\sqrt{2}

    I = \displaystyle 6\sqrt{2}

    is this correct

    OR

    \displaystyle \lim_{\epsilon \to 2}\left[   -6\sqrt{2-\epsilon}   \right]_0^{\epsilon}
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  2. #2
    MHF Contributor
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    I don't understand how you got a factor of 6... This is solved using a \displaystyle u substitution...
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by bijosn View Post
    I = \displaystyle \int_0^{2} \frac{1}{\sqrt{2-x}} dx

    I = \displaystyle -6\sqrt{2-x} \right]_0^{2}

    I = \displaystyle -6\sqrt{0} + 6\sqrt{2}

    I = \displaystyle 6\sqrt{2}

    is this correct
    No. You can make sure by differentiating the anti-derivative.
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  4. #4
    Junior Member bijosn's Avatar
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    Quote Originally Posted by Prove It View Post
    I don't understand how you got a factor of 6... This is solved using a \displaystyle u substitution...
    sorry, I forgot to add the 3 before the integral sogn in the first post...now fixed

    I = \displaystyle 3 \int_0^{2} \frac{1}{\sqrt{2-x}} dx
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  5. #5
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    Then yes, it is correct.
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