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Integral problem reverse order

  1. Nov 6, 2008 #1
    Integral problem "reverse order"

    This is from an online homework question:

    Evaluate the integral by reversing the order of integration.

    [tex]\int^{1}_{0}\int^{8}_{8y} e^{x^{2}}dxdy[/tex]

    Although, I fail to see how this works, if I switch the order I get:

    [tex]\int^{8}_{8y}\int^{1}_{0} e^{x^{2}}dydx[/tex]

    after integrating wrt y, it boils down to

    [tex]\int^{8}_{8y} e^{x^{2}}dy[/tex]

    which won't give me a number... did I do something wrong?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 6, 2008 #2


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    Re: Integral problem "reverse order"

    The limits of your integral depend on y, so you can't just switch the order of integration without changing the limits accordingly. Your integration region is {y: [0,1], x: [8y,8]}, so to switch the order of integration you need to figure out what the limits of y are in terms of x and what the constant limits of x are.

    It helps to draw a picture of the region of integration. y runs from 0 to 1 along the y-axis, and since x is from 8y to 8, it runs from 0 to 8 along the x-axis, but you're only considering the area below the line x = 8y, or equivalently y = x/8.

    So, the limits could equivalently be written as x: [0,8], y:[x/8,1]. So, when you go to integrate over y, you need to switch to these limits when you integrate.
  4. Nov 6, 2008 #3
    Re: Integral problem "reverse order"

    Ohhh, thanks Mute. My bad.

    So, after integrating wrt y, it becomes

    [tex]\int^{8}_{0}[e^{x^{2}} - \frac{x}{8}e^{x^{2}}] dx[/tex]

    requiring integration by parts, with a substitution...etc. Yes?
  5. Nov 6, 2008 #4


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    Re: Integral problem "reverse order"

    The outer integral is from 0 to 1/8, not from 0 to 8.
    [tex]\int xe^{x^2} dx[/tex]
    can be integrated by the substitution u= x2.

    [itex]\int e^{x^2} dx[/tex[
    does not have any elementary function as it anti-derivative.
  6. Nov 6, 2008 #5
    Re: Integral problem "reverse order"

    So, your saying y runs from 0 to 1/8 and x runs from 1 to 1/8?

    that doesn't make sense to me...
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