Question

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

Answer 1

$f\left(x\right)=\{\begin{array}{cc}x-\left[x\right]& \text{if}\left[x\right]\text{is odd},\\ 1+\left[x\right]-x& \text{if}\left[x\right]\text{is even},\end{array}$

$f\left(x\right)$ and $\cos \pi x$ are both periodic with period $2$ and are both even.
$\therefore \underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx=2\underset{0}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx$
$=10\underset{0}{\overset{2}{\int }}f\left(x\right)\cos \pi xdx$
$\underset{0}{\overset{1}{\int }}f\left(x\right)\cos \pi xdx=\underset{0}{\overset{1}{\int }}(1-x)\cos \pi xdx=-\underset{0}{\overset{1}{\int }}u\cos \pi udu$
$\underset{1}{\overset{2}{\int }}f\left(x\right)\cos \pi xdx=\underset{1}{\overset{2}{\int }}(x-1)\cos \pi xdx=-\underset{0}{\overset{1}{\int }}u\cos \pi udu$
$\therefore \underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx=-20\underset{0}{\overset{1}{\int }}u\cos \pi udu$
$=\frac{40}{{\pi }^2}$
$\Rightarrow \frac{{\pi }^2}{10}\underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx=4$