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Double integration of an exp function

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data



    Please help I need to find the area of e^(-x^2) with the domain 0 is less than or equal to y which is less than or equal to 1 and y is less than or equal to x which is less than or equal to one

    2. Relevant equations



    3. The attempt at a solution
    I set up the equation ∫ ∫ e^(-x^2) dydx with values y=x and y=0 and x=1 and x=0
    attempted to integrate but struggling.
     
  2. jcsd
  3. Mar 22, 2012 #2
    Let's see your attempt at integration then
     
  4. Mar 22, 2012 #3
    Well I started by integrating with respect to y and got ye^(-x^2)dx,I then solved for the values y=x and y=0 leaving xe^(-x^2)dx and unsure of if my first steps are correct and how to continue.
    My idea for a next step would be integration by substitution u=-x^2 but unsure as to whether that would work
     
  5. Mar 22, 2012 #4
    why -x^2?
    I'd just go for x^2

    Try it and see what happens
     
  6. Mar 22, 2012 #5
    Ok here goes u= x^2 that gives ∫ e^(-u) integration by substitution gives me -(e^x^2)/2
    Possibly :)

    Now do I just solve for x=1 and x=y?
     
  7. Mar 22, 2012 #6
    When you did your first integration, you essentially integrated the strip from 0 to y=x, in the second integration you are taking all of those strips from 0 to 1, you don't need to solve for x=1 or x=y, you just need to evaluate the integral from 0 to 1
     
  8. Mar 22, 2012 #7
    Sorry when I said solve I meant evaluate.
    Could I switch the order of integration so when I evaluate after integrating a second time I lose x and y from the equation?
     
  9. Mar 22, 2012 #8
    Try it and see what happens
    You'll have trouble evaluating it if you do the dx integration first however, the integral of e^(-x^2)dx isn't nice, it isn't solvable in terms of elementary functions. If you're interested, the name of the solution is the error function.
     
  10. Mar 22, 2012 #9
    Thanks for your help will just leave it now and not confuse myself even more
     
  11. Mar 22, 2012 #10
    The best way to get use to it is to just play about with it and see what you end up with.
     
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