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Improper Integral

  1. Nov 2, 2007 #1
    For the integral from 0 to infinity of


    I am getting as far as:

    -1/5*x*e^(-5x) + 1/5*int of (e^(-5x)dx)

    But I am getting stuck at this point. We are supposed to come out with 1/25 for the answer but how would I evaluate the "-1/5*x*e^(-5x)" since that is already out of the integral????

    Thank you in advance.
  2. jcsd
  3. Nov 2, 2007 #2


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  4. Nov 2, 2007 #3
    yes do it by parts
  5. Nov 3, 2007 #4
    does the integral really converge to 1/25?
  6. Nov 3, 2007 #5


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    Yes, 1/25 is the answer.
  7. Nov 3, 2007 #6
    How does it converge to 1/25? Do you have to evaluate the the result as a limit by applying the L'hopital rule? Or is there a better way of evaluating this integral?
  8. Nov 3, 2007 #7
    No, forget about the deadly L'hopital's rule, you only apply that on something that gives you 0/0 or infinity/infinity.

    Your basic method of plugging in the limits of your integral, after you get the solution works here. Do you know what e^(-infinity) turns out to be?. That will simplify your solution.
  9. Nov 3, 2007 #8


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    Are you saying that you cannot evaluate xe-5x at 0 and [itex]\infty[/itex]? That's easy! at 0, you have 0*1= 0 and any polynomial time e-x goes to 0 as x goes to [itex]\infty[/itex] so that part is 0.

    Of course, the integral of 1/5 e-5x is -1/25 e-5x. At x= [itex]\infty[/itex] that is 0 and at x= 0, it is -1/25. The difference is 1/25
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