Proving Average Value of Gaussian Function as k_0

[thenewbosco]

The question says show that the average value of k is k_0 for the function A(k)=\frac{1}{\sqrt{2\pi}a}e^{\frac{-(k-k_0)^2}{2a^2}.

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?

 

[HallsofIvy]

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