Then the value of $${{{\pi ^2}} \over {10}}\int\limits_{ - 10}^{10} {f\left( x \right)\cos \,\pi x\,dx} $$ is
Given,
$$f(x) = \left\{ {\matrix{ {x - [x]} & {if\,[x]\,is\,odd} \cr {1 + [x] - x} & {if\,[x]\,is\,even} \cr } } \right.$$
f(x) and cos $$\theta$$ x both are periodic with period 2 and both are even.
$$\therefore$$ $$\int\limits_{ - 10}^{10} {f(x)\cos \pi x\,dx} $$ $$ = 2\int\limits_0^{10} {f(x)\cos \pi x\,dx} $$
$$ = 10\int\limits_0^2 {f(x)\cos \pi x\,dx} $$
Now, $$\int\limits_0^1 {f(x)\cos \pi x\,dx} $$
$$ = \int\limits_0^1 {(1 - x)\cos \pi x\,dx = - \int\limits_0^1 {u\cos \pi u\,du} } $$ and $$\int\limits_1^2 {f(x)\cos \pi x\,dx} $$
$$ = \int\limits_1^2 {(x - 1)\cos \pi x\,dx = - \int\limits_0^1 {u\cos \pi u\,du} } $$
$$\therefore$$ $$\int\limits_{ - 10}^{10} {f(x)\cos \pi x\,dx} $$
$$ = - 20\int\limits_0^1 {u\cos \pi u\,du = {{40} \over {{\pi ^2}}}} $$
$$ \Rightarrow {{{\pi ^2}} \over {10}}\int\limits_{ - 10}^{10} {f(x)\cos \pi x\,dx = 4} $$
