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Integration by parts

  1. Aug 14, 2010 #1
    Integration by parts - Exponential distribution

    1. The problem statement, all variables and given/known data
    Solve the following definite integral:

    [tex]\int^{\infty}_{0} \frac{1}{\lambda} x e^{-\frac{x}{\lambda}} dx[/tex]

    I'm asked to solve this integral. The solution is [tex]\lambda[/tex], although I'm not sure how this was done.

    2. Relevant equations

    3. The attempt at a solution
    [tex]\int^{\infty}_{0} \frac{1}{\lambda} x e^{-\frac{x}{\lambda}} dx[/tex]

    [tex]= \frac{1}{\lambda} \int^{\infty}_{0} x e^{-\frac{x}{\lambda}} dx[/tex]

    [tex]=\frac{1}{\lambda} \left( \left[ x e^{-\frac{x}{\lambda}} \right] ^{\infty}_{0} - \int^{\infty}_{0} e^{-\frac{x}{\lambda}} dx \right) [/tex], integration by parts.

    The [tex] \left[ x e^{-\frac{x}{\lambda}} \right] ^{\infty}_{0} [/tex] term, by fundamental theorem of calculus is 0. Thus,

    [tex]= - \int^{\infty}_{0} e^{-\frac{x}{\lambda}} dx \right) [/tex],

    I don't know what to do at this point, because as far as I know, taking the definite integral of this term will result in [tex]e^{-\frac{x}{\lambda}}[/tex] , which, solving for 0 and infinity will yield -1.

    Where have I gone wrong?

    I appreciate your input.

    Last edited: Aug 14, 2010
  2. jcsd
  3. Aug 14, 2010 #2
    Close. You forgot the chain rule. The derivative of e^-u is -e^-u. (or antiderivative.)
  4. Aug 14, 2010 #3
    Thanks whitish,

    I was in the middle of editing the formula after your post. Do you mind looking at what I have posted again?

  5. Aug 15, 2010 #4
    nevermind. I see whats happening now, and i'm getting the same answer as you are. with your last integral you can just use U substitution. are you sure just lambda is the right answer?
    Last edited: Aug 15, 2010
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