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Average value gaussian

  1. Nov 16, 2006 #1
    The question says show that the average value of k is [tex]k_0[/tex] for the function [tex]A(k)=\frac{1}{\sqrt{2\pi}a}e^{\frac{-(k-k_0)^2}{2a^2}[/tex].

    I know that this is a gaussian function and that the center is k0 but i am not sure how to show this is the average value of k, what equation for average value would i evaluate. is it some kind of integral? and are the limits + and - infinity?
     
  2. jcsd
  3. Nov 17, 2006 #2

    HallsofIvy

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    For any probability distribution, f(x), the "average" value (more correctly, the mean value), also called the "expected" value, is given by
    [tex]\int xf(x)dx[/itex]. While [itex]e^{x^2}[/itex] does not have an elementary anti-derivative, putting that additional x in makes it easy to integrate.
     
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