For any real number x , let x denote the largest integer less than or equal to x . Let f be a real-valued…

$f\left(x\right)=\left\{\begin{array}{l}x-1,\\ 1-x,\end{array}\begin{array}{l}1\le x<2\\ 0\le x<1\end{array}\right.$

$f\left(x\right)$ is periodic with period $2$, and is even.

$\therefore I={\int }_{-10}^{10} f\left(x\right) \mathrm{cos\pi }x dx$

$= 2{\int }_0^{10}f\left(x\right) \mathrm{cos\pi }x dx$

$=2\times 5{\int }_0^{2}f\left(x\right) \mathrm{cos\pi }x dx$

$=10 \left[{\int }_0^1\left(1-x\right) \mathrm{cos\pi }x dx+{\int }_1^2\left(x-1\right) \mathrm{cos\pi }x dx\right]=10\left(I_1+I_2\right)$

$I_2={\int }_1^2\left(x-1\right) \mathrm{cos\pi }x dx$

Put $\left(x-1\right)=t$

$I_2=-{\int }_0^{1}t\mathrm{cos\pi }tdt$

$I_1={\int }_0^1\left(1-x\right) \mathrm{cos\pi }x dx=-{\int }_0^{1}x\cos \left(\pi x\right) dx$

$\therefore I=10\left[-2{\int }_0^{1}x\mathrm{cos\pi }x dx\right]=-20{\left[x\frac{\mathrm{sin\pi }x}{\mathrm{\pi }}+\frac{\mathrm{cos\pi }x}{{\mathrm{\pi }}^2}\right]}_0^1$

$=-20\left[-\frac{1}{{\mathrm{\pi }}^2}-\frac{1}{{\mathrm{\pi }}^2}\right]=\frac{40}{{\mathrm{\pi }}^2}$

$\therefore \frac{{\mathrm{\pi }}^2}{10}I=4$