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Double integration using trig term

  1. Jul 22, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^1_0 \int^1_y sin(x^2) dx dy[/tex]

    3. The attempt at a solution

    This equation cannot be integrated nicely, so I tried to reverse the order of integration:

    [tex]\int^1_0 \int^1_x sin(x^2) dy dx[/tex]

    However this only helps for the first step, since when we intregrate by y, we get:

    [tex]\int^1_0 sin(x^2)-xsin(x^2) dx[/tex]

    I'm stuck... again! Any ideas?
     
  2. jcsd
  3. Jul 22, 2008 #2

    Defennder

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    Homework Helper

    Your limits in the new double integral is incorrect. If in doubt, look at the original double integral and draw a picture of the region enclosed by the limits. Then try to express the limits of the double integral in the reversed order of the same region.
     
  4. Jul 22, 2008 #3
    Alright so perhaps the bounds are then: [tex]\int^1_0 \int^x_0[/tex]

    In which case, my integral simplifies to:

    [tex]\int^1_0 x sin(x^2)[/tex]

    Which becomes:

    -1/2 cos (x^2) | 1_0

    This make sense?
     
  5. Jul 22, 2008 #4

    Defennder

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    Homework Helper

    Yep, you got it.
     
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