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Math Help - Calculus : U-Substitution

  1. #1
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    Calculus : U-Substitution

    The problem is x(16-3x)^1/2 dx

    u = 16 - 3x
    du = -3 dx


    My problem is how do I get rid of the X in the original problem? I can divide -1/3 to get rid of the -3 and that'll take care of the dx but as for the X I have no idea how to go about it. Help =)
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  2. #2
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    Quote Originally Posted by Afterme View Post
    The problem is x(16-3x)^1/2 dx

    u = 16 - 3x
    du = -3 dx


    My problem is how do I get rid of the X in the original problem? I can divide -1/3 to get rid of the -3 and that'll take care of the dx but as for the X I have no idea how to go about it. Help =)
    u = 16-3x

    3x = 16-u

    x = \frac{1}{3}(16-u)

    substitute ...

    \int \frac{1}{3}(16-u) \cdot u^{\frac{1}{2}} \cdot \frac{du}{-3}

    -\frac{1}{9} \int 16u^{\frac{1}{2}} - u^{\frac{3}{2}} \, du
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  3. #3
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    Quote Originally Posted by skeeter View Post
    u = 16-3x

    3x = 16-u

    x = \frac{1}{3}(16-u)

    substitute ...

    \int \frac{1}{3}(16-u) \cdot u^{\frac{1}{2}} \cdot \frac{du}{-3}

    -\frac{1}{9} \int 16u^{\frac{1}{2}} - u^{\frac{3}{2}} \, du

    Thanks alot =)
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  4. #4
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    distributive property of multiplication ...

    (16 - u)u^{\frac{1}{2}} = 16 \cdot u^{\frac{1}{2}} - u \cdot u^{\frac{1}{2}} = 16u^{\frac{1}{2}} - u^{\frac{3}{2}}
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