2.2.13 * Given the function $f(x)=4 \pi \arcsin x-[\arccos (-x)]^{2}$ with the maximum value $M$ and the minimum value $m$. Then $M-m=$ $\qquad$
Given that $\arccos (-x)=\pi-\arccos x$, $\arcsin x=\frac{\pi}{2}-\arccos x$, therefore
$$
\begin{aligned}
f(x) & =4 \pi\left(\frac{\pi}{2}-\arccos x\right)-(\pi-\arccos x)^{2} \\
& =-(\arccos )^{2}-2 \pi \arccos x+\pi^{2} \\
& =-(\arccos x+\pi)^{2}+2 \pi^{2},
\end{aligned}
$$

so $f_{\max }(x)-\pi^{2}, f_{\min }(x)=-2 \pi^{2}$. Hence $M-m=3 \pi^{2}$.