Mathematics - Definite Integration Question with Solution | TestHub

Given, $f(x)=\{\begin{array}{cc}x-[x]& \text{ if }[x]\text{ is odd }\\ 1+[x]-x& \text{ if }[x]\text{ is even }\end{array}$ $f(x)$ and $\cos \pi x$ both are periodic with period 2 and both are even. $$\begin{array}{rl}\therefore & {\int }_{-10}^{10}f(x)\cos \pi xdx\\ & =2{\int }_0^{10}f(x)\cos \pi xdx\end{array}$$
$$\begin{array}{rl}& =10{\int }_0^{2}f(x)\cos \pi xdx\\ & \text{ Now, }{\int }_0^{1}f(x)\cos \pi xdx\\ & ={\int }_0^1(1-x)\cos \pi xdx=-{\int }_0^{1}u\cos \pi udu\\ & \text{ and }{\int }_1^{2}f(x)\cos \pi xdx\\ & ={\int }_1^2(x-1)\cos \pi xdx=-{\int }_0^{1}u\cos \pi udu\\ & \therefore {\int }_{-10}^{10}f(x)\cos \pi xdx\\ & =-20{\int }_0^{1}u\cos \pi udu=\frac{40}{{\pi }^2}\\ & \Rightarrow \frac{{\pi }^2}{10}{\int }_{-10}^{10}f(x)\cos \pi xdx=4\end{array}$$