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Math Help - Tricky integration

  1. #1
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    Tricky integration

    \int_0^2 r^3  \sqrt{4-r^2}  dr

    Not sure how to integrate this... I tried a u substitution for 4-r^2 but couldn't find anything to replace r^3 with.
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Tricky integration

    try sub u=4-r^2
    so r^2 = 4-u
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  3. #3
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    Re: Tricky integration

    Quote Originally Posted by deezy View Post
    \int_0^2 r^3  \sqrt{4-r^2}  dr

    Not sure how to integrate this... I tried a u substitution for 4-r^2 but couldn't find anything to replace r^3 with.
    \displaystyle \begin{align*} \int{r^3\sqrt{4 - r^2}\,dr} = -\frac{1}{2}\int{-2r\cdot r^2\sqrt{4 - r^2}\,dr} \end{align*}

    Let \displaystyle \begin{align*} u = 4 - r^2 \implies du = -2r\,dr \end{align*} and the integral becomes

    \displaystyle \begin{align*} -\frac{1}{2}\int{-2r\cdot r^2\sqrt{4 - r^2}\,dr} &= -\frac{1}{2}\int{\left(4 - u\right)\sqrt{u}\,du} \\ &= -\frac{1}{2}\int{4u^{\frac{1}{2}} - u^{\frac{3}{2}}\,du} \\ &= -\frac{1}{2}\left(\frac{4u^{\frac{3}{2}}}{\frac{3}{  2}} - \frac{u^{\frac{5}{2}}}{\frac{5}{2}}\right) + C \end{align*}

    Now after you change the terminals to u terminals and insert them, you will have the answer for the definite integral.
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  4. #4
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    Re: Tricky integration

    You can also do

    u = r^2
    du = 2rdr
    dr = \frac{1}{2}du

    so \frac{1}{2}\int (4-u)udu
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  5. #5
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    Re: Tricky integration

    Quote Originally Posted by Intrusion View Post
    You can also do

    u = r^2
    du = 2rdr
    dr = \frac{1}{2}du

    so \frac{1}{2}\int (4-u)udu
    Where did the square root go?
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