Technon said:
The expectation value of any function ##f(x)## is given by
In QM, expectation values aren't properties of functions; they're properties of operators. The expectation value of an operator ##\hat{A}## is (note that we're using a particular representation, the one-dimensional position representation, in all of this--things look different in other representations)
$$
\langle \hat{A} \rangle = \int_{- \infty}^{\infty} \psi^* (x) \hat{A} \psi (x) dx
$$
For example, the position operator ##\hat{Q}## is just ##x##, i.e., "multiply by ##x##"; so its expectation value is
$$
\langle \hat{Q} \rangle = \int_{- \infty}^{\infty} x \psi^* (x) \psi (x) dx
$$
which is similar to what you wrote (note that we can rearrange the factors inside the integral because multiplication of functions is commutative). Similarly, an operator that was something like "multiply by ##f(x)##", where ##f(x)## is some function of ##x##, would have an expectation value that looks similar to what you wrote.
However, this won't be true for all operators, because not all operators can be represented as "multiply by some function ##f(x)##". For example, the momentum operator ##\hat{P}## is ##- i \hbar d / dx##, so its expectation value is
$$
\langle \hat{P} \rangle = - i \hbar \int_{- \infty}^{\infty} \psi^* (x) \frac{d \psi (x)}{dx} dx
$$
which is
not similar to what you wrote.
Technon said:
what do these functions represent on a physical level?
As should be evident from the above, the functions represent operators, but, as noted, not all operators can be represented as functions.