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Integrating with U-substitution

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the integral of (1-x)((2x-x^2)^.5) dx


    2. Relevant equations



    3. The attempt at a solution
    I am kind of hazy on U substitutions but I thought that was the right way to go here:
    let u=2x-x^2
    du=2-2x
    (1/2) Integral of (u)^.5 du < i wasn't sure if I went wrong here, Since i had (1-x) in the original equation and my du=2(1-x) can i put du in my substitution as long as I put (1/2) to balance the equation?

    using this I got (2x-x^2)^(3/2)/(6) + C which was marked wrong. any insight would be appreciated
     
  2. jcsd
  3. Sep 14, 2009 #2
    Your original integral is:

    [tex]\int{\frac{1-x}{\sqrt{2x-x^2}}dx[/tex]

    You did the substitution well.

    [tex]u=2x-x^2[/tex]

    [tex]du=(2-2x)dx=2(1-x)dx[/tex]

    So [itex]dx=du/2(1-x)[/itex]

    Now just substitute back in the integral (for dx and u)

    [tex]\int{\frac{1-x}{\sqrt{u}}*\frac{du}{2(1-x)}}[/tex]
     
  4. Sep 14, 2009 #3
    oh no the initial integral is (1-x)Squareroot(2x-x^2) all in the numerator
     
  5. Sep 14, 2009 #4
    [tex]
    \int{{1-x}{sqrt{2x-x^2}}dx
    [/tex]
     
  6. Sep 14, 2009 #5

    tiny-tim

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    Hi zooboodoo! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    Yes, you've done everything perfectly, except you got the factor, 6, wrong at the end …

    differentiate what you have (using the chain rule), and you'll see. :smile:
     
  7. Sep 14, 2009 #6
    wah it was supposed to be /3 wasn't it :-x so frustraating
     
  8. Sep 14, 2009 #7

    tiny-tim

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    yeah … it's so easy to make a mistake like that! :frown:

    always check any integration problem by differentiating your answer, and making sure you get back to where you started! :wink:
     
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