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Double integral question

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Sketch the region of integration and then evaluate the double integral:


    2. Relevant equations

    [tex]\int\int[/tex]x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

    3. The attempt at a solution

    I have managed to do half of the problem and integrate it respect to x but then have no idea how to finish the problem.

    I mangaged to get upto [tex]\int[/tex]ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

    can someone please tell me how to finish it or if i have done something wrong so far??

    cheers
     
    Last edited: Apr 10, 2009
  2. jcsd
  3. Apr 10, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi kieranl! Welcome to PF! :smile:

    (have a ≤ :wink:)
    uhh? :confused:

    start again, and integrate over dy first! :smile:
     
  4. Apr 12, 2009 #3
    but if i integrated by dy first wouldnt i end up with an answer containing y's and not numbers??
     
  5. Apr 12, 2009 #4

    tiny-tim

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    If you integrate with respect to dy, then you eliminate y, and only have x's.
     
  6. Apr 12, 2009 #5
    but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?
     
  7. Apr 12, 2009 #6
    Draw a picture of the region R. Then use the picture to write R in the form

    R={(x,y):a<x<b, f(x)<y<g(x)}.
     
  8. Apr 13, 2009 #7

    tiny-tim

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    Hi kieranl! :smile:

    (have an integral: ∫ and a ≤ :wink:)
    No, you have to integrate over 0 ≤ y ≤ x ≤ 1 …

    whether you do it over x first or over y first is up to you. :wink:
     
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