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Double integral e^(ysqrtx)dxdy

  1. Apr 24, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫∫e^(y√x)dxdy from 1 to 4 then from 0 to 2

    2. Relevant equations

    ∫ e^x = e^x
    u substitution

    3. The attempt at a solution

    I am just curious if this is equal to double integral e^(y\sqrt{x})dydx from 0 to 2 then from 1 to 4. In other words can I change the order of integration without screwing up my function? If so I can solve it. If not I have tried U substitution:

    u = y√x
    du = y/2√x} dx

    which changes my equation to double integral (2√x/y)e^(u)dudy from 1 to 4 then from 0 to 2 which is equal to (2u/y^2)e^(u)dudy from 1 to 4 then from 0 to 2 but I don't see how integrating that will give me 2e^u(u-1)/y^2 which was wolfram alpha's indefinite integration for e^y\sqrt{x}. help?
     
  2. jcsd
  3. Apr 24, 2014 #2

    SammyS

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    Assuming you integral is :
    [itex]\displaystyle \int_0^2\int_1^4 {e^{y\sqrt{x}}}\,dx\,dy[/itex]​
    It should be fine to switch the order of integration.

    If you're trying to do the integration in the given order, then remember you need to treat y as a constant as you integrate with respect to x. Remember then to evaluate that as a definite integral over x.
     
  4. Apr 24, 2014 #3
    thanks! that's what I thought, so now I can solve it
     
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