Answer: Since $f(x)=9-x^2$ is an even function. Hence in the Fourier series $b_n=0.$..............................(1)

Fourier series expansion of any function is given as:

$f(x)=\dfrac {a_0}{2} +\sum_{n=1}^\infty a_n cos(\dfrac{n \pi x}{l})+\sum_{n=1}^\infty b_n sin(\dfrac{n \pi x}{l}),$

where, $a_n= \dfrac{1}{l} \int_{-l}^l f(x)cos(\dfrac{n \pi x}{l})dx,$ $a_0= \dfrac{1}{l} \int_{-l}^l f(x)dx,$ $b_n= \dfrac{1}{l} \int_{-l}^l f(x)sin(\dfrac{n \pi x}{l})dx.$

Here $l=3.$ Now $a_0= \dfrac{1}{3} \int_{-l}^l (9-x^2)dx= \dfrac{2}{3} \int_0 ^3(9-x^2)dx=\dfrac{2}{3}[9x-\dfrac{x^3}{3}]_0^3=\dfrac{2}{3}[27-9]=12. $...................(2)

$a_n= \dfrac{1}{3} \int_{-3}^3 f(x)cos(\dfrac{n \pi x}{3})dx=\dfrac{1}{3} \int_{-3}^3 (9-x^2)cos(\dfrac{n \pi x}{3})dx$

As it is an even function, hence

$=\dfrac{2}{3} \int_{0}^3 (9-x^2)cos(\dfrac{n \pi x}{3})dx$

$=\dfrac{2}{3} \Bigg[(9-x^2)\bigg (\dfrac{3}{n \pi}sin\bigg(\dfrac{n \pi x}{3}\bigg)\bigg)-(-2x)\bigg (\dfrac{-9}{n^2 \pi^2}cos\bigg(\dfrac{n\pi x}{3}\bigg)\bigg)+(-2)\bigg (\dfrac{-27}{n^3 \pi^3}sin\bigg(\dfrac{n \pi x}{3}\bigg)\bigg)\Bigg]_0^3$

First and Third term of above expression is zero, hence only second term remains.

$=\dfrac{2}{3} \Bigg [-\dfrac{18 \times 3}{n^2 \pi^2} \Bigg]cos n \pi= -\dfrac{36}{n^2 \pi^2}(-1)^n=a_n$........................................................................................(3)

Hence Fourier series expansion is $f(x)=9-x^2=a_0 +\sum_{n=1}^\infty a_n cos(\dfrac{n \pi x}{3})+\sum_{n=1}^\infty b_n sin(\dfrac{n \pi x}{3}).$

Substituting $a_0, a_n$ and $b_n$ form (1), (2) and (3), we get

$f(x)=9-x^2=12-\dfrac{36}{\pi^2} \sum_{n=1}^\infty(-1)^n \dfrac{cos(n\pi x/3)}{n^2}.$                                     Answer.