Given the following question:
The function $f(x)$ is defined by $$ f(x)= \begin{cases} 2x& \textrm{if} \ 0 \leq x<\dfrac{\pi}{3} \\ \pi - x & \textrm{if} \ \dfrac{\pi}{3}\leq x \leq \pi \end{cases} $$
Also given the following info:
$f(-x)=-f(x) = f(x+2\pi)$
I know that the graph is a triangle wave form and it repeats every 2 period. I also know it is odd and to use $$b_n=\frac{1}{L}\int_{-L}^L f(x) \sin \left(\frac{n\pi}{L}x\right)\ dx$$ The problem I have is I don't know how to find the $f(x)$ value. Any help would be appreciated.
I don't want to solve your problem, but try to give you some hints.
Your function is a pice-wise defined function, so you musn't expect an expression like $f(x)=2x+3/2$, but an expression like
$$ f(x)=\begin{cases} 2x, & \mbox{when } 0\leq x<\pi/3 \\ \pi-x & \mbox{when } \pi/3\leq x<\pi . \end{cases} $$
On the other hand, they tell you your function is odd ($f(-x)=-f(x)$), so you can achive a similar expresion but when $x$ is in $[-\pi,0]$ (try it).
Your total function will be the ''paste'' of two data, and is defined only on $[-L,L]$, $L=\pi$. The Fourier series extends it to the whole real line.
Now, how do yo compute coefficients? Well, since the function is odd you know that the only coefficients of $\sin$ matters. All you have to do it now is divide your integral in ''little'' integrals where you have an expression like $x^2$ for you function. I mean, cut the intergal interval $[-L,L]$ in some short intervals.
Finally, note that the argument is an even function (is the product of two odd) and is integred in a symmetric interval. Then you can reduce your above to expression to
$$ b_n=\frac{2}{L}\int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) dx. $$
PD: To get the symbol $\pi$ you must type
\pi. Similar, for $\leq$ or $\geq$ you have to type\leq(Less or EQual) and\geq(Great or EQual), respectively.