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A very simple U-sub is wrecking my brain

  1. Jul 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Integral Sqrt[16-x^2]*x dx from 0 to 2.

    2. Relevant equations



    3. The attempt at a solution

    First thing I do is a u-sub, so

    u=16-x^2

    du=-2xdx

    I also my bounds change to upper=12 lower=16.

    After I integrate and swap out my U, I get

    -1/3*(16-x^2)^(3/2) from 16 to 12.

    Am I using the wrong method?
     
  2. jcsd
  3. Jul 19, 2009 #2
    This is probably a stupid question, but why did you change the limits of integration?
     
  4. Jul 19, 2009 #3

    Nabeshin

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    Umm, this is actually a trig-substitution problem (which I guess is a class of u-substitution).

    Try thinking of it in those terms.
     
  5. Jul 19, 2009 #4
    Have you thought of trying a trig substitution?
     
  6. Jul 19, 2009 #5
    You sure it's necessary? I get an answer that matches my calculator without incorporating any trig.

    EDIT: Oh.
     
  7. Jul 19, 2009 #6
    You could do the problem using a u substitution, it's just easier to use the trig substitution.
     
  8. Jul 19, 2009 #7

    Nabeshin

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    I actually missed the *x in the problem the first glance through, but I still would go with trig substitution. Necessary, maybe not.
     
  9. Jul 19, 2009 #8
    Are my bounds incorrect? I thought you had to adjust your bounds when using a usub.
     
  10. Jul 19, 2009 #9

    Nabeshin

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    You adjust if you keep the equation in terms of u, but if you substitute back in for x after integration, you retain the original x boundaries. Seems you substituted back into x and changed the boundaries, which doesn't work.
     
  11. Jul 19, 2009 #10
    This clears up a lot. Thanks. I got the correct answer now!
     
  12. Jul 19, 2009 #11

    HallsofIvy

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    That is one reason why it is a good idea to write the variable in the limits of integration:
    [tex]\int_{x= 0}^2 \sqrt{16- x^2} xdx[/tex]
    after the substitution [itex]u= 16- x^2[/itex] becomes
    [tex]-\frac{1}{2}\int_{u=16}^{12}u^{1/2}du[/tex]
     
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