For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real-valued function defined on the interval [-10,10]…

f(x)={\begin{matrix}x-1, \\ 1-x,\end{matrix}\begin{matrix}1\leq x<2 \\ 0\leq x<1\end{matrix}

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

$$\therefore I={\int }_{-10}^{10}f(x)cos\pi xdx$$∴I=∫−1010​f(x)cosπxdx

$$=2{\int }_{0}^{10}f(x)cos\pi xdx$$=2∫010​f(x)cosπxdx

$$=2\times 5{\int }_{0}^{2}f(x)cos\pi xdx$$=2×5∫02​f(x)cosπxdx

$$=10[{\int }_{0}^{1}(1-x)cos\pi xdx+{\int }_{1}^{2}(x-1)cos\pi xdx]=10({I}_{1}+{I}_{2})$$=10[∫01​(1−x)cosπxdx+∫12​(x−1)cosπxdx]=10(I1​+I2​)

$${I}_{2}={\int }_{1}^{2}(x-1)cos\pi xdx$$I2​=∫12​(x−1)cosπxdx

Put $(x-1)=t$(x−1)=t

$${I}_{2}=-{\int }_{0}^{1}tcos\pi tdt$$I2​=−∫01​tcosπtdt

$${I}_{1}={\int }_{0}^{1}(1-x)cos\pi xdx=-{\int }_{0}^{1}x\mathrm{cos}(\pi x)dx$$I1​=∫01​(1−x)cosπxdx=−∫01​xcos(πx)dx

$$\therefore I=10[-2{\int }_{0}^{1}xcos\pi xdx]=-20{[x\frac{sin\pi x}{\pi }+\frac{cos\pi x}{{\pi }^{2}}]}_{0}^{1}$$∴I=10[−2∫01​xcosπxdx]=−20[xπsinπx​+π2cosπx​]01​

$$=-20[-\frac{1}{{\pi }^{2}}-\frac{1}{{\pi }^{2}}]=\frac{40}{{\pi }^{2}}$$=−20[−π21​−π21​]=π240​

$$\therefore \frac{{\pi }^{2}}{10}I=4$$∴10π2​I=4