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Math Help - Using Substitution with Integrals

  1. #1
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    Using Substitution with Integrals

    Can someone please show me how to do this problem? I have the answer but do not understand the steps in order to arrive at it.

    Integral 4r square root(6-r) dr



    Answer: 8/5*(6-r)^5/2 - 16(6-r)^3/2 +C
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  2. #2
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    Do you know integration by parts?

    If not, read Integration by Parts -- from Wolfram MathWorld.

    If so, use it.
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  3. #3
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    I read that but still can't get it to work out properly.
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  4. #4
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    4\int r\sqrt{6-r} \, dr

    let u = 6-r

    r = 6-u

    dr = -du

    substitute ...

    4 \int (u-6)\sqrt{u} \, du

    4 \int u^{\frac{3}{2}} - 6u^{\frac{1}{2}} \, du

    can you finish?
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  5. #5
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    ok, but how do you get the 8/5 and the -16?
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  6. #6
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    4 \int u^{\frac{3}{2}} - 6u^{\frac{1}{2}} \, du

    antiderivative ...

     4\left[\frac{2}{5}u^{\frac{5}{2}} - 4u^{\frac{3}{2}}\right] + C

    distribute the constant 4 ...

    \frac{8}{5}u^{\frac{5}{2}} - 16u^{\frac{3}{2}} + C

    now back substitute (6-r) for u.
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