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Double Integral Problem

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Evaluate [tex]\int[/tex][tex]\int[/tex] e^x^2 dx dy.

    The bounds for the inner integral go from y to 1
    The bounds for the outer integral go from 0 to 1



    2. The attempt at a solution

    I can easily do this, I just do not see how I can get e^x^2 to integrate for x. Is there some sort of special method to do this?
     
  2. jcsd
  3. Apr 2, 2008 #2

    Dick

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    Do you mean (e^x)^2 or e^(x^2)? The answer is quite different depending.
     
  4. Apr 2, 2008 #3
    Honestly, I can't tell. Its a take home test and the subscript goes:

    e^x^2. Looks to me like its e^(x^2) so that's how I'm going to solve it.
     
  5. Apr 2, 2008 #4

    Dick

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    In that case it's trouble. e^(x^2) doesn't have an elementary antiderivative. (e^x)^2=e^(2x) does. Somebody should pay better attention to putting parentheses into problems to clarify them.
     
  6. Apr 2, 2008 #5
    Alright well I guess I'll just have to email my professor and ask him to clarify it.
     
  7. Apr 5, 2008 #6
    Okay so I talked to my professor and you can actually do this problem. It has something to do with switching the boundaries. Like instead of having it dxdy, make it dydx and change the boundaries of the integrals using the graph of the region.

    Since the region is a right triangle with the hypotenuse going from (0,0) to (1,1), I believe that the boundaries go as follows: 0 < x < 1 and x<y<1 where the < are less than greater than.

    So that would make the problem go as follows:

    Integrating e^x^2 in terms of y gives (y*e^x^2 ). Then plugging in boundaries is where I get stuck again. I'm not totally sure if my new boundaries are right.
     
  8. Apr 5, 2008 #7

    Dick

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    Write either e^(x^2) or (e^x)^2 depending on what you mean, ok? e^x^2 doesn't mean anything. Sure you can interchange integration order etc. The only problem with doing it is if you interpret the integrand as e^(x^2). I thought that was what you were going to clarify with your professor?
     
  9. Apr 5, 2008 #8
    f(x, y) = ex2 is constant in y. Switching the order of integration gives us the integral:
    [tex]\int_0^1\int_0^x e^{x^2} dy dx[/tex]
    This simplifies easily, as you grab an x from the limits of integration. The resulting integral is elementary.
     
    Last edited: Apr 5, 2008
  10. Apr 5, 2008 #9
    To be honest seeing as it's very easy to work out an answer if it's

    1:[tex]\int_0^1\int_y^1 (e^x)^2\; dx[/tex]

    And it's more difficult but still possible to evaluate

    2:[tex]\int_0^1\int_y^1 e^{x^2}\; dx[/tex]

    I'd say it's most likely the second one. But that said as said above we need to know which is which.

    So which is it eq. 1: or eq. 2:?

    Indeed slider this is true, it's actually not that awkward, if you know what you are doing either way. However it becomes much more difficult when you don't know what the question is. :wink::smile:
     
  11. Apr 5, 2008 #10

    Dick

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    That's an excellent point that I completely missed. Thanks.
     
  12. Apr 5, 2008 #11
    But then doesn't the x*e^x^2 integrated over dx bring me back to the same problem of being unable to integrate e^x^2?
     
  13. Apr 5, 2008 #12

    Dick

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    No, do a substitution. u=x^2, du=2xdx. And STOP writing e^x^2, that's not clear.
     
  14. Apr 6, 2008 #13
    hint: click on equation 2 from my post above then cut and paste the latex into the post window. Then everyone will know what you mean.

    [tex]\int_0^1\int_0^x e^{x^2} dy dx[/tex]

    [tex] \int xe^u\;dx[/tex]
     
    Last edited: Apr 6, 2008
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