Given f(x)=4-x2 in interval (0,2),however the standard interval is (0,2l)

Hence we get l=1.

Let f(x)=a0 + $\sum$an cos $\dfrac{n\pi x}{l} $ + $\sum $bn  sin $\dfrac{n\pi x}{l} $

$\therefore$ f(x)=a0 + $\sum $an $\cos(n\pi x)$   +$\sum$bn $\sin(n\pi x)$.........(1)

a0=$\dfrac{1}{2l}\int_0^{2l} {f(x)}\,\mathrm{d}x$=$\dfrac{1}{2}\int_0^{2} {(4-x^2)}\,\mathrm{d}x$

$\therefore$a0=$\dfrac{1}{2}$$\bigg[4x-\dfrac{x^3}{3}\bigg]_0^2$=$\dfrac{1}{2}\bigg[8-\dfrac{8}{3}\bigg]$=$\dfrac{8}{3}$

   an=$\dfrac{1}{l}\int_0^{2l}{f(x)} .\cos\bigg(\dfrac{n\pi x}{l}\bigg)\,\mathrm{d}x$

$\therefore $an=$\int_0^{2}{(4-x^2)} .\cos({n\pi x})\,\mathrm{d}x$

$\therefore $an=$\bigg[(4-x^2)\bigg (\dfrac{\sin( n\pi x)}{n\pi}\bigg)$-$(-2x)\bigg(-\dfrac{\cos(n\pi x)}{n^2\pi^2}\bigg)$+$(-2)\bigg(-\dfrac{\sin(n\pi x)}{n^3}\bigg)$$\bigg]_0^2$ $\therefore $**an=$\bigg[{0-\dfrac{4}{n^2\pi^2}+0}-{0-0-0}\bigg]$=$-\dfrac{4}{n^2\pi^2}$**

   bn=$\dfrac{1}{l}\int_0^{2l} {f(x)}.\sin\bigg(\dfrac{n\pi x}{l}\bigg)\,\mathrm{d}x$

$\therefore$bn=$\int_0^{2} {(4-x^2)}.sin({n\pi x})\,\mathrm{d}x$

$\therefore$bn=$\bigg[(4-x^2)\bigg(-\dfrac{\cos(n\pi x)}{n\pi}\bigg)-(-2x)\bigg(-\dfrac{\sin(n\pi x)}{n^2\pi^2}\bigg)+(-2)\bigg(\dfrac{\cos(n\pi x)}{n^3\pi^3}\bigg)\bigg]_0^2$

$\therefore$bn=$\bigg[\{0-0-\dfrac{2}{n^3\pi^3}\}-\{-\dfrac{4}{n\pi}-0-\dfrac{2}{n^3\pi^3}\}\bigg]=\dfrac{4}{n\pi}$

$\therefore$f(x)=x2-4=$\dfrac{8}{3}-\dfrac{4}{\pi^2}\bigg[\dfrac{1}{1^2}\cos(\pi x)+\dfrac{1}{2^2}\cos(2\pi x)+\dfrac{1}{3^2}\cos(3\pi x)+.....\bigg]-\dfrac{4}{\pi}\bigg[\dfrac{1}{1}\sin(\pi x)+\dfrac{1}{2}\sin(2\pi x)+\dfrac{1}{3}\sin(3\pi x)+.....\bigg]$...Substituting values of a0,an,bn in (1)

• For deduction part we substitute x=0 in equaion above

$\therefore $$4-0=\dfrac{8}{3}-\dfrac{4}{\pi^2}\bigg[\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....\bigg]$ $\therefore $$\dfrac{1}{3}=-\dfrac{1}{\pi^2}\bigg[\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....\bigg]$.................(i)  and when x=2 we have,     $4-4=\dfrac{8}{3}-\dfrac{4}{\pi^2}\bigg[\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+......\bigg]$ $\therefore $$-\dfrac{2}{3}=-\dfrac{1}{\pi^2}\bigg[\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+......\bigg]$.............(ii) * Adding (i) and (ii) we get , $-\dfrac{1}{3}=-\dfrac{2}{\pi^2}\bigg[\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.......\bigg] $ $\therefore $ $\dfrac{\pi^2}{6}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....$