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Definite double integral of e^x^2

  1. Oct 9, 2012 #1
    Hi guise. I just encountered a problem which I sincerely don't know how to attack. I don't know what kind of variable substitution would help me to solve this problem... It's that goddamn e^x^2 which is a part of the integrand... I don't know if I should use polar coordinates either... Please help!

    [itex]\int_{1}^{2} (\int_{x}^{x^3}x^2 e^{y^2}dy)dx + \int_{2}^{8} (\int_{x}^{8}x^2 e^{y^2}dy)dx[/itex]

    P.s. The answer is supposedly equal to (1/6)(62e^64+e).
     
    Last edited: Oct 9, 2012
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  3. Oct 9, 2012 #2

    dextercioby

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    I don't see a method of solution (if there exists one), but I can tell you for sure it's incorrect to compute the integral wrt y in <polar coordinates>.
     
  4. Oct 9, 2012 #3

    Hurkyl

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    I'm not a costume. :tongue:

    The obvious things to try, then, would be [itex]u = x^2[/itex] and [itex]u = e^{x^2}[/itex].

    Another common trick is to swap the order of integration and hope you get something simpler out.
     
  5. Oct 9, 2012 #4
    I tried the substitution u= x^2, v= e^(y^2). The Jacobian had the value 1/(4xy*e^(y^2)). So far so good. But then I get...
    [itex]\frac{1}{4} \int_{1}^{4} (\int_{e^u}^{e^{u^3}}(\frac{u^{1/2}}{ln(v)^{1/2}})dv)du +\frac{1}{4} \int_{2}^{8^{2}} (\int_{e^u}^{e^{8^2}}(\frac{u^{1/2}}{ln(v)^{1/2}})dv)du[/itex]

    Which is a complete**#@* mess.


    And integrating with respect to x first would not be sufficient to eliminate all expressions containing x from the integrand of neither the first nor second term.
     
    Last edited: Oct 9, 2012
  6. Oct 9, 2012 #5
    I was just thinking one thing... Let the set of points (region, whatever... you get the idea) which the first term is to be integrated over be denoted A.

    Then A = {(x,y): x ≤ y ≤ x^3, 1 ≤ x ≤ 2}. Let the set of points that the second term is integrated over be denoted B.
    Then B = {(x,y): x ≤ y≤ 8, 2 ≤ x ≤ 8}.

    It becomes clear that these two regions of points in the xy-plane intersect at the line x=2. Is it possible to... somehow... reexpress the boundaries for y and x in a way which JUST MAYBE makes the integration easier? Is it possible to "merge" the two terms into one term, with modified integration intervals?

    I don't know people... but this problem is supposed to have a solution without the necessity of fancy mathematical methods... but those are welcome as well! As long as they are thoroughly explained. <3

    I tried the substitution u=x^2 and v= y^2, but that didn't help either... I always get to the goddamn errorfunction which I basically don't know anything about. This **** is supposed to be solved with elementary functions. Not the error function...
     
  7. Oct 10, 2012 #6

    HallsofIvy

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    Try changing the order of integration. The first integral has y going from x to x3 and x from 1 to 2. So swapping the order of integration would give
    [tex]\int_{y= 1}^8\int_{x= y^{1/3}}^y x^2e^{y^2}dxdy= \frac{1}{3}\int_1^8 (y^3- y)e^{y^2}dy[/tex]

    Since we now have odd powers of y, we can write that as
    [tex]\frac{1}{3}\int_1^8 (y^2- 1)e^{y^2}(ydy)[/tex]
    and use the substitution [itex]u= y^2[/itex] so that du= 2ydy and the integral becomes
    [tex]\frac{1}{6}\int_1^{64} (u- 1)e^u du[/tex]
     
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