For any real number \(x\), let \([x]\) denotes the largest integer less than or equal to \(x\). Let \(f\) be a real valued function defined on the interval \([-10,10]\) by \(f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.\).
Then the value of \(\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x\) is

Given, \(f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.\) \(f(x)\) and \(\cos \pi x\) both are periodic with period 2 and both are even.
\[
\begin{aligned}
\therefore \quad & \int_{-10}^{10} f(x) \cos \pi x d x \\
& =2 \int_0^{10} f(x) \cos \pi x d x
\end{aligned}
\]


\[
\begin{aligned}
& =10 \int_0^2 f(x) \cos \pi x d x \\
& \text { Now, } \quad \int_0^1 f(x) \cos \pi x d x \\
& =\int_0^1(1-x) \cos \pi x d x=-\int_0^1 u \cos \pi u d u \\
& \text { and } \int_1^2 f(x) \cos \pi x d x \\
& =\int_1^2(x-1) \cos \pi x d x=-\int_0^1 u \cos \pi u d u \\
& \therefore \quad \int_{-10}^{10} f(x) \cos \pi x d x \\
& =-20 \int_0^1 u \cos \pi u d u=\frac{40}{\pi^2} \\
& \Rightarrow \quad \frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x=4
\end{aligned}
\]