Starting with
$$
\langle x\rangle=\int \Psi^{*}(x, t) x \Psi(x, t) d x
$$
and the time-dependent Schrödinger equation, show that
$$
\frac{d\langle x\rangle}{d t}=\int \Psi^{*} \frac{i}{\hbar}(\hat{H} x-x \hat{H}) \Psi d x
$$
Given that
$$
\hat{H}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x)
$$
show that
$$
\hat{H} x-x \hat{H}=-2 \frac{\hbar^{2}}{2 m} \frac{d}{d x}=-\frac{\hbar^{2}}{m} \frac{i}{\hbar} \hat{P}_{x}=-\frac{i \hbar}{m} \hat{P}_{x}
$$
Finally, substitute this result into the equation for $d\langle x\rangle / d t$ to show that
$$
m \frac{d\langle x\rangle}{d t}=\left\langle\hat{P}_{x}\right\rangle
$$
Interpret this result.