Average value of $e^{-x}$ from $0$ to $\infty$ I just signed up for this site so I hope I'm doing this right.
I've been having trouble wrapping my head around the average value of the function $e^{-x}$ from 0 to $\infty$. I know that the average value of a function is:

$$\bar f =\frac{\int_a^b{f(x)dx}}{\int_a^b{dx}}=\frac{\int_a^b{f(x)dx}}{b-a}$$  

And when a is 0 and b is $\infty$:

$$\frac{\int_0^\infty{f(x)dx}}{\int_0^\infty{dx}}$$

And taking the limit since the integral is improper:

$$\lim_{b\to\infty}\frac{\int_0^b{f(x)dx}}{\int_0^b{dx}}$$

When I evaluate the average value of the function $e^{-x}$ I get a result of zero (one over infinity) by taking the following steps:

$$\lim_{b\to\infty}\frac{\int_0^b{e^{-x}dx}}{\int_0^b{dx}}=\lim_{b\to\infty}\frac{-e^{-x}|_0^b}{x|_0^b}=\lim_{b\to\infty}\frac{-e^{-b}+1}{b}=0$$

While this makes some mathematical sense to me (Unless I made a mistake in which case please let me know!) I'm puzzled on how the average value of this function can be zero when all defined values of the same function are greater than zero. Does this conclusion make any sense or am I off-track entirely?

I find it intuitive. We start with a value of $1=f(0)$ and decrease to $0$ as $n \to \infty$. If we find $f(n)$ for $n=0,1,2,3,...$ The more terms we choose, the more terms we have close to zero, and thus our average gets closer to $0$. Just think about it if you have $100$ numbers close to $0$ despite that $1$ in the begging isn't our average going to be close to $0$?. Although what I'm describing is more of an approximation with rectangles (width $=1$ and thus its height $f(n)$ is the area), the same idea applies if you use $f(n)$ $n=0,0+h,0+2h,...$
