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Double integrals with IBP.

  1. Jun 27, 2014 #1
    1. The problem statement, all variables and given/known data
    ∫∫[ye^(-xy)]dA R=[0,2]×[0,3] evaluate the integral.


    2. Relevant equations



    3. The attempt at a solution
    So I started with some algebra changing the integral to ∫(e^-x)[∫ye^-ydy]dx
    I evaluated the y portion first because its more difficult to deal with and wanted to get it out of the way.
    I ended up integrating by parts with:
    U=y dv=e^-x
    Dy=dy v=-e^-x and got
    -ye^(-y) - ∫-e^(-y)dy on the interval [0,3] and got -4e^(-3) + 1

    This is now a constant, pulled it out of the x integral leaving:
    (-4e^(-3) +1)∫e^(-x)dx
    The final integral I evaluated as:
    -e^(-x) on [0,2] gives
    -e^(-2) + e^0 and this is multiplied by the previous number to give:
    4e^(-6)-4e^(-3)-e^(-2)+1
    Using a calculator to approximate I get a value of 0.692468
    The answer in the book is .5e^(-6) + (5\2) which is approximated by a calculator as 2.5012394

    I'm doing something wrong and I am wondering if there is a rule I forgot, do I need to integrate by x first and then y?
     
  2. jcsd
  3. Jun 27, 2014 #2

    ehild

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    It is wrong: ##(e^{-x})(ye^{-y})=ye^{-x-y}## instead of ##ye^{-xy}##

    ehild
     
  4. Jun 27, 2014 #3
    I was walking back from the corner store when I realized this, but thank you for confirming it. I had, as another helper once told me "abused the chain rule."
     
  5. Jun 27, 2014 #4

    LCKurtz

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    Oh my! Chain rule abuse! Oh Guard -- Guard!! Come quickly before he gets away!
     
  6. Jun 28, 2014 #5

    verty

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    Hint: if y was constant, the integrand would be quite simple.
     
  7. Jun 28, 2014 #6

    HallsofIvy

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    Verty's hint suggest that you do the integration in the other order:
    [tex]\int_{y=0}^3\left[\int_{x= 0}^2 ye^{-xy}dx\right] dy[/tex]

    You should find that much simpler.
     
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