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Math Help - tricky integral

  1. #1
    niz
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    tricky integral

    Hi all,
    Here's a integral problem which has been baffling me. Please help me find its solution.

    2 3 3
    ∫ [x √(8-x ) ] dx
    0

    ( Sorry I am not good at writing this integral clear enough. In words it's Integral of x and cube root of (8 - x cube) dx with integral range from 0 to 2. Hope this helps.)
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by niz View Post
    Hi all,
    Here's a integral problem which has been baffling me. Please help me find its solution.

    2 3 3
    ∫ [x √(8-x ) ] dx
    0

    ( Sorry I am not good at writing this integral clear enough. In words it's Integral of x and cube root of (8 - x cube) dx with integral range from 0 to 2. Hope this helps.)
    Thanks in advance.
    So it's \int_0^2{x\sqrt[3]{8 - x^3}\,dx}?
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  3. #3
    niz
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    Quote Originally Posted by Prove It View Post
    So it's \int_0^2{x\sqrt[3]{8 - x^3}\,dx}?
    Yes that's the integral. Thanks
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     I \int_0^2 x  \sqrt[3]{8-x^3}~dx

    By substituting  x = 2 \sqrt[3]{t}

     dx = \frac{2}{3} t^{-\frac{2}{3}} dt

    we have  I = \int_0^1 2 \sqrt[3]{t} \cdot  2 \sqrt[3]{1 -t } ~ \cdot \frac{2}{3} t^{-\frac{-2}{3}} dt

     = \frac{8}{3} \int_0^1 t^{ \frac{1}{3} - \frac{2}{3} } (1 - t)^{\frac{1}{3}}

     = \frac{8}{3} \int_0^1 t^{-\frac{1}{3}} (1 - t)^{\frac{1}{3}}

     = \frac{8}{3} B( \frac{2}{3} , \frac{4}{3} )

     = \frac{16\pi}{9\sqrt{3}}
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  5. #5
    niz
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    Hi Simplependulum,
    Thanks a lot for your help but I am bit unclear about the 2nd last step which consists of a function B(2/3, 4/3). What's this function? Can you please throw some light on it? Thanks once again.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by niz View Post
    Hi Simplependulum,
    Thanks a lot for your help but I am bit unclear about the 2nd last step which consists of a function B(2/3, 4/3). What's this function? Can you please throw some light on it? Thanks once again.
    Beta Function -- from Wolfram MathWorld
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  7. #7
    niz
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    Thanks a lot.
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