f(x) = |x|

$\therefore f(-x)= |-x|= |x| = f(x)$

$\therefore f(x) \ $is even function

$\therefore b_n = 0$

Here, $l=\pi$

Now, $a_0 = \dfrac{2}{l} \displaystyle \int^1_0 f(x) dx$

$= \dfrac {2}{\pi} \displaystyle \int^\pi _0 |x| dx$

$\displaystyle =\dfrac {2}{\pi} \int^\pi_0 xdx$

$= \dfrac {2}{\pi}\left [ \dfrac {x^2}{2} \right ]^\pi_0$ …