Let $f(x)$ be the function on [−3, 3] which is graphed below,
Find the constant term in the Fourier series for $f$?
$$\frac{a_{0}}{2}=\frac{1}{2L}\int^{L}_{-L}{f(x)}dx=\frac{1}{2\times3}\int^{3}_{-3}{f(x)}dx=\frac{1}{2\times3}\bigg[\int^{0}_{-3}{f(x)}dx+\int^{3}_{0}{f(x)}dx\bigg]$$
From here on, I have no idea where to go
Before doing calculations, it is useful to grasp their interpretation and get intuition on the result.
The constant term in the Fourier series is simply the average value of the function over one period. Here, your function est antisymmetric around $(0,2.5)$. So the respective areas above and below $2.5$ compensate. I guess $5/2$ could be the desired result. You now perform the actual computations. You can compute integrals, and you can do geometry: draw vertical lines below discontinuities $(-3,-1,0,1,3)$, and sum areas of trapezoids.
All results should match.
A graphical explanation: for the average
The area between the black and the blue curve are the same, except for the sign. So the area of the blue curve is the same as that of the black curve, which is easy to compute.
If you really want to compute areas with trapezoids, here are four of them. The first on the left has $2\times 1$ (bottom rectangle) plus $ 2\times 2/2$ (top triangle. Finally, from left to right, you add four trapezoids and you get $4+1.5+3.5+6=15$. The average value is thus $15/(3-(-3))=2.5$.