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Double Integral of e^-y

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^x_0\int^y_x e^{-v} dv du[/tex]

    where u and v are just "dummy variables"

    2. Relevant equations



    3. The attempt at a solution

    [tex]\int^x_0\int^y_x e^{-v} dv du = \int^x_0 -e^{-y} + e^{-x} du = (-e^{-y} + e^{-x})x[/tex]

    have I made a mistake somewhere?
     
  2. jcsd
  3. Mar 27, 2009 #2

    djeitnstine

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    Everything looks fine don't forget your integration constant after ;)
     
  4. Mar 27, 2009 #3
    It's for determining a CDF. I don't think I need the constant.

    A classmate got x(1 - e-y). Did they use somesort of trick to get it, or is their answer just wrong?
     
  5. Mar 27, 2009 #4

    djeitnstine

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    That's very wrong. First of all there has to be at least 2 exponentials, perhaps they just forgot their [tex]e^{-x}[/tex]
     
  6. Mar 27, 2009 #5
    Well, that cleared that up.

    Thanks.
     
  7. Mar 27, 2009 #6
    Got another question.

    [tex]\int^x_0\int^y_x e^{-v} dv du [/tex] This case is when 0<x<y

    If I want to do 0 < y < x, would the double integral be set up in the following way:

    [tex]\int^x_0\int^x_y e^{-v} dv du [/tex] ?
     
  8. Mar 27, 2009 #7

    djeitnstine

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    Hmm before I go into any discussion, are you integrating backwards?? Or are you trying to change the order of integration? If you are changing the order I would say this:

    You're changing the order of integration. Lets say your limits are [tex]y=x^2[/tex] from 0 < x 6 bounded by y=0 and you are doing the dy dx order or integration then you would have to do
    [tex]\int^{6}_{0} \int^{x^2}_{0} dy dx[/tex] (from top to bottom, left to right)

    Now say you want to change the order, that is, from dx dy, you would have to write y in terms of x and do your integration that way, i.e.

    [tex]\int^{36}_{0} \int^{6}_{\sqrt{x}} dx dy[/tex] (from left to right, bottom to top)

    if not just ignore it and let me know what you mean
     
  9. Mar 27, 2009 #8
    I'm trying to find the CDF of e^-y.

    The cdf of any f(x,y) is [tex]\int^x_{-\infty} \int^y_{-\infty} f(u,v) dv du[/tex]

    where u and v are just dummy variables. I need to consider 2 cases: 1) 0<x<y and 2) 0 < y < x. In other words I need to do 2 double integrals using the general set up of a CDF.
     
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