${\int }_{-\pi }^{\pi }\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|\mathrm{d}x$

Let $f\left(x\right)=\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|$

Replace $x$ by $-x$.

$f\left(-x\right)=\left|\mathrm{\pi }-\left|-\mathrm{x}\right|\right|\Rightarrow f\left(-x\right)=\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|$

Since $f\left(x\right)=f\left(-x\right) \Rightarrow f\left(x\right)$ is an even function.

Now by the property of definite integration for even function.

${\int }_{-\pi }^{\pi }\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|\mathrm{d}x=2{\int }_0^{\pi }\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|\mathrm{d}x$

$=2{\int }_0^{\pi }\left|\mathrm{\pi }-\mathrm{x}\right|\mathrm{d}x \left(\because x>0\right)$

$=2{\int }_0^{\pi }\left(\mathrm{\pi }-\mathrm{x}\right)\mathrm{d}x \left(\because \mathrm{\pi }-x>0 \forall x\in \left(0, \mathrm{\pi }\right)\right)$

$=2{\overline{)\left(\mathrm{\pi x}-\frac{{\mathrm{x}}^2}{2}\right)}}_0^{\mathrm{\pi }}$

$=2\left({\pi }^2-\frac{{\mathrm{\pi }}^2}{2}-0-0\right)={\mathrm{\pi }}^2$