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Line integral answer

  1. Apr 14, 2007 #1
    Int ((2xe^y)dx + (x^2e^y) dy) from (0,0) to (1,-1)

    I get the answer 2/e, while my book says 1/e. Am I right or wrong?
     
  2. jcsd
  3. Apr 14, 2007 #2

    nrqed

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    I agree with your book. If you show your work, I might be able to point out your mistake. (by the way, it's because the integral of x gives x^2/2 that I "lose" the factor of 2)
     
  4. Apr 14, 2007 #3
    2e^y*INT(x)dx + x^2*INT(e^y)dy

    = e^y*x^2 + x^2*e^y = 2e^y*x^2

    = 2*e^(-1) =2/e
     
  5. Apr 15, 2007 #4

    nrqed

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    This is not the way to do it!! You have to choose a path and integrate along that path. For example, go from the origin to (1,0) and integrate your expression (note that for that path, dy=0). Then go from (1,0) to (1,-1) and integrate again (this time, dx will be zero). Add the two results.
     
  6. Apr 15, 2007 #5
    OK, then I made it. But why is the method of my first attempt wrong? I integrate along a path there as well; the straigth line between the endpoints.
     
    Last edited: Apr 15, 2007
  7. Apr 15, 2007 #6

    HallsofIvy

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    An "Anti-derivative" of 2xeydx+ x2eydy (which is an exact differential) is x2e^y. Evaluating that at (0,0) gives 0 and at (1, -1) gives e-1. The integral is e-1= 1/e.

    As for the way you did it, you did NOT integrate along the straight line between the endpoints. The line from (0,0) to (1, -1) is given by parametric equations x= t, y= -t so dx= dt and dy= -dt. The integral becomes
    [tex]\int_0^1 2te^{-t}dt- \int_0^1t^2e^{-t}dt= \int_0^1(t- t^2)e^{-t}dt[/tex]
    Using integration by parts should give you the same answer. It appears that all you did was integrate 2xeydx with respect to x while ignoring the y and integrate x2eydy with respect to y while ignoring the x- you can't do that. In effect you got the "2" because you integrated the same thing twice!
     
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