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Improper Integral

  1. Mar 27, 2008 #1
    Improper Integral [Solved]

    1. The problem statement, all variables and given/known data

    [tex]
    \int_{0}^{\infty} (x-1)e^{-x}dx
    [/tex]

    2. Relevant equations

    Integration by Parts
    Improper Integrals


    3. The attempt at a solution

    [tex]
    \lim_{R\rightarrow \infty} \int_0^R~xe^{-x}-e^{-x}dx
    [/tex]

    Let u = x
    du = dx

    Let dv = e^-x
    v = -e^-x

    [tex]
    -xe^{-x} - \int -e^{-x}dx
    [/tex]

    [tex]
    -xe^{-x}-e^{-x} - \int e^{-x}dx
    [/tex]

    [tex]
    = -xe^{-x}
    [/tex]

    [tex]
    \lim_{R\rightarrow \infty}-xe^{-x} \mid_{0}^{R}=0
    [/tex]

    Now there is clearly area when I look at the graph, but the graph intersects the x-axis at x=1 not x=0 I’m wondering if this is some sort of trick with the bounds of integration that I don’t see or if I made a mistake with my integration.

    Thanks
     
    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No integration mistake. But remember that the 'area' an integral measures is area above the x-axis minus area below the x-axis. In your case these areas are equal.
     
  4. Mar 27, 2008 #3
    Thanks.

    I guess since the area wasn’t symmetrical I just assumed it could not be zero, and that I had made a mistake.
     
  5. Mar 27, 2008 #4

    dynamicsolo

    User Avatar
    Homework Helper

    When you graph [tex]xe^{-x}-e^{-x}[/tex] , it does seem remarkable that that little sliver above the x-axis from x = 1 out to infinity is actually going to cancel out that curved wedge between x = 0 and x = 1. But you can satisfy yourself that you've made no mistake in your integration by evaluating the areas from x = 0 to x=1 and from x = 1 to x -> infinity separately. The areas are -1/e and +1/e , so you're OK...
     
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