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Integration question

  1. Feb 23, 2005 #1
    Could anyone evaluate this integral for me? I got it in an exam and didn't know how to do it.

    [tex]\int\sqrt{x-x^2}[/tex]
     
    Last edited: Feb 23, 2005
  2. jcsd
  3. Feb 23, 2005 #2
    let u = x-1/2 and do the substitude,...
     
  4. Feb 23, 2005 #3
    What do you mean? Can you evaluate it completely showing the steps?
     
  5. Feb 23, 2005 #4
    Ok, Mathematica was definitely wrong. There's going to be a trig substitution in this integral somewhere.
     
    Last edited: Feb 23, 2005
  6. Feb 23, 2005 #5
    Mathematica is **** and I'm sure thats not a right answer.
     
  7. Feb 23, 2005 #6
    what do I mean? I thought I make it very clear already!
    [tex]\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx[/tex]

    let [tex] u=x-\frac{1}{2} [/tex]

    It become:

    [tex]\int\sqrt{(\frac{1}{2}-u)(\frac{1}{2}+u)}du[/tex]

    [tex]=\int\sqrt{\frac{1}{4}-u^2 }du[/tex]

    [tex]=\frac{u}{4}\sqrt{1-4u^2}+\frac{1}{8}sin^{-1}(2u)+C[/tex]

    [tex]=\frac{x-1/2}{4}\sqrt{1-4(x-1/2)^2}+\frac{1}{8}sin^{-1}(2x-1)+C[/tex]
     
  8. Feb 24, 2005 #7
    Sick, thanks.
     
  9. Feb 25, 2005 #8

    dextercioby

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    And one more thing:"Mathematica" is ALWAYS RIGHT...:grumpy:

    Daniel.
     
  10. Feb 25, 2005 #9

    HallsofIvy

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    Cute factoring.

    What I would have done is "complete the square": x- x2= 1/4-1/4+ x-x2= 1/4-(x-1/2)2 which would have led me to the u= x- 1/2 substitution. Once I had it in the form [itex]\sqrt{\frac{1}{4}- u^2}[/itex] I would use a trig substitution.
     
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