I have a function like $f(x)$ and $x$ has a probability distribution form $0$ to $x_\text{max}$ like $p(x)$.
Intuitively, my measurements (in the Lab) shows that the integration of $f(x)$ over $p(x)$ yields the same results as $f(x_\text{mean})$. Is there any mathematical proof in general form for this? Or at least in what conditions the integration of a function over its distribution can be approximated with the value of the function at the mean point?
In general if $X$ admits density $p$, $$\mathbb E[f(X)]=\int_\mathbb R f(x)p(x)dx$$ and $$\mathbb E[X]=\int_\mathbb R x p(x)dx$$ and there are even cases where $$ f(\mathbb E[X])=0,\ \mathbb E[f(X)]=+\infty$$ however, if $p(x)$ is uniform on an interval $[0,b]$ you can do it in the case were $$f(x)=\alpha x+g(x)+o(x-\frac{b}{2})$$ the sum of a linear function, a function which is $g$ is odd over $[0,b]$ and a term that is near zero in the interval. Indeed you have $$ \mathbb E[f(X)]=\int_0^b \alpha x \ dx+\int_0^b g(x) \ dx+\int_0^b o(x-\frac{b}{2}) dx=$$ Since $g(\cdot)$ is odd over $[0,b]$ and $o(x-\frac{b}{2})$ is small $$\sim \int_0^b \alpha x \ dx=\alpha \mathbb E[X]\sim f(\mathbb E[X])$$ since $g(\frac{b}{2})=0$