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Iterated Integral

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following double integral. Change order of integration if necessary.

    [tex]\int^{1}_{0} \int^{x}_{0} x^2 sin(\Pi x y) dy dx[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]\int^{1}_{0} \int^{x}_{0} x^2 sin(\Pi x y) dy dx = -\frac{1}{\Pi}\int^{1}_{0} x cos(\Pi x^2 ) dx[/tex]

    Let u = x^2 and du = 2x dx

    [tex] - \frac{1}{2 \Pi} \int^{1}_{0} cos (\Pi u) du = -\frac{1}{2 \Pi} \frac{sin (\Pi x^2 )}{\Pi} |^{1}_{0} = - \frac{sin( \Pi)}{2 \Pi^2} = 0[/tex]

    but that's wrong. Anyone catch my mistake?

    I was also wondering when I'm supposed to change the order of integration. Thanks.
     
    Last edited: Feb 8, 2009
  2. jcsd
  3. Feb 8, 2009 #2
    Hint: What is the value of [tex]\cos(0)[/tex]?
     
  4. Feb 8, 2009 #3
    cos(0) = 1

    I see what you meant, let me try it
     
    Last edited: Feb 8, 2009
  5. Feb 8, 2009 #4
    I was referring to when you integrated the sine with respect to y, and received a cosine. The lower integration limit is 0 so it cos(0) should not vanish.



    Edit: I see you discovered what I meant. It took 20+ minutes for this computer to load my reply!
     
  6. Feb 8, 2009 #5
    Took me a little while, but I got it. Thank you.

    I was just wondering when the right time to change the order of integration is, since we never covered it in class.
     
  7. Feb 9, 2009 #6

    HallsofIvy

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    When it is convenient. Certainly if you can't do a double integral in a given order you should try changing the order.
     
  8. Feb 9, 2009 #7
    That certainly makes sense. Thank You.
     
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