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Homework Help: Where is the mistake?

  1. Jul 17, 2007 #1
    (Double integral on D) sin[((x^3)/3) - x] dxdy = ? where
    D={(x,y): 1<=y<=4 , sqrt(y)<=x<=2 }.

    Okay, we change the order of integration and then we get

    (Double integral on D') sin [((x^3)/3) - x] dxdy where
    D'={(x,y): 1<=x<=2 , 1<=y<=(x^2). Thus, we get the one variable integral

    (Integral from x=1 to x=2) (x^2 - 1)* sin[((x^3)/3) - x]dx.

    Letting u = ((x^3)/3) converts the integral to

    (Integral from u=-2/3 to u=2/3) sin(u)du = 0. So where is the mistake?
    Last edited: Jul 17, 2007
  2. jcsd
  3. Jul 17, 2007 #2


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    Isn't this correct?
  4. Jul 17, 2007 #3
    Shouldn't the substitution be u = x^3/3 - x? Which would make the bounds in u space be -2/3 to 5/3?

    Edit: Nevermind, -2/3 to 2/3 is the correct bounds.
    Last edited: Jul 17, 2007
  5. Jul 17, 2007 #4
    Pardon, the substitution is of course u = x^3/3 - x but i have found the integral 0. The answer in the worksheet is different though.
    Last edited: Jul 17, 2007
  6. Jul 17, 2007 #5
    Woops... I see no problem with your work, and it looks like the integral should indeed be 0.
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