Finding expectation value of $p^2$ without integrals So the expectation value of momentum, if you know the expectation value of position is 
$$\langle p \rangle = m \frac{d\langle x \rangle}{dt}$$
Is there a nice formula like this for $\langle p^2 \rangle$? 
$$p^2 = m^2v^2 $$
$$\langle p^2 \rangle = m^2\langle v^2 \rangle$$
$$\langle p^2 \rangle = m^2\Big\langle \frac{dx}{dt}\frac{dx}{dt} \Big\rangle ?$$
 A: There's a problem of notation here. If you are assuming that $\langle \cdot \rangle$ is the expectation in time, ie: $\langle \cdot \rangle = \int_0^\infty \cdot dt$ then $\frac{d \langle x \rangle}{dt} =0$ by definition. The expectation does not change in time because it's the time-averaged value! 
However, let's assume that you are using an ensemble average or a spatial average or something else to avoid this issue. 
In this case, the expression you gave is perfectly valid to find the expectation of $p^2$. However, a note of caution -- you said that you know the expected value of position. This does not mean you know $\langle \frac{dx}{dt} \frac{dx}{dt} \rangle$. 
More specifically, 
$$\bigg\langle \frac{dx}{dt} \frac{dx}{dt} \bigg\rangle \neq \frac{d\langle x \rangle}{dt}\frac{d\langle x \rangle}{dt}$$
Also note that your equations are assuming mass is constant, which is probably valid but it's worth noting why the $m$ is outside the expectation. If mass were not constant, you'd have the same issue I noted about the expectation of the position. 
