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Integral involving exponential logarithm

  1. Aug 21, 2010 #1
    1. The problem statement, all variables and given/known data
    If X is a random variable with density function: f(x) = [tex]\lambda[/tex]e[tex]^{-x \lambda}[/tex]where X>=0.


    2. Relevant equations
    Why is the expected value of X, or E[X] = [tex]\frac{1}{\lambda}[/tex]?


    3. The attempt at a solution
    E[X] = [tex]\int[/tex] x*([tex]\lambda[/tex]e[tex]^{- \lambda}[/tex][tex]^{x}[/tex]) dx, where the integral is from 0 to infinity.

    I let u = -[tex]\lambda[/tex]x
    du = -[tex]\lambda[/tex] dx

    but I can't get the [tex]\frac{1}{\lambda}[/tex] as the answer when I performed the integration.
     
  2. jcsd
  3. Aug 21, 2010 #2

    HallsofIvy

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    If you let [itex]u= \lambda x[/itex] then [itex]x= u/\lambda[/itex] and the integral becomes
    [tex]\int \frac{u}{\lambda}e^{-u} du= \frac{1}{\lambda}\int ue^{-u}du[/tex]

    Did you remember to replace the "x" multiplying the exponential with [itex]u/\lambda[/itex]?
     
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