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Homework Help: 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.

    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!

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

    It become:


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


  8. Feb 24, 2005 #7
    Sick, thanks.
  9. Feb 25, 2005 #8


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

  10. Feb 25, 2005 #9


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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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