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Double Integral of exp(y)/y

  1. Apr 19, 2012 #1
    Suppose the question is:

    ∫∫exp(y)/y dydx

    Now the first integral is w.r.t. y and goes from 0 to 1. The second integral is that of x and goes form y to y^2.

    I've evaluated it multiple times and it comes out to be 2-e. ( I first integrate the integrand w.r.t. x then w.r.t. y as the other way I believe its not possible). Anyway for some reason 2-e is not the correct answer. Am I doing something wrong?
     
  2. jcsd
  3. Apr 19, 2012 #2

    [itex]\int_0^1\int_y^{y^2} \frac{e^y}{y}\,dxdy=\int_0^1\frac{e^y}{y}(y^2-y)\,dy=\int_0^1 ye^y\,dy-\int_0^1e^y\,dy=ye^y|_0^1-2\int_0^1 e^y\,dy=2-e[/itex]

    I may be wrong, but I think that either whoever/whatever is telling you the result is wrong is wrong, or you're not doing the appropriate exercise. Check this.

    DonAntonio
     
  4. Apr 19, 2012 #3
    I'm glad my answer was correct. Thanks DonAntonio.
     
  5. Apr 19, 2012 #4

    Hurkyl

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    Where did you get the integral? I speculate that you were given a region to integrate over, but you got the bounds wrong.

    Notice that 2-e is negative, but your integrand is always positive. This is actually correct for the problem you stated (can you see why?), but it strongly hints at the mistake I'm guessing at.
     
  6. Apr 19, 2012 #5

    HallsofIvy

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    Hurkyl's point is that y^2 is less than y for y between 0 and 1 so that your integral is from the higher number to the lower. It is more common (though not "mathematically" required) for the lower limit to be less than the upper limit. If you are integrating over a region in the plane then that will be necessary. Reversing the limits of integration will change the sign so you would get e- 2 (a positive number) rather than 2- e, a negative number.
     
  7. Apr 19, 2012 #6
    Turns out that while the professor intended it to be y^2 to y (and so the ans would be e-2) he mistakenly wrote it the other way around. Thanks for clearing the confusion.
     
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