Odd / Even integrals My textbook doesn't really have an explanation for this so could someone explain this too me. 
If f(x) is even, then what can we say about: $$\int_{-2}^{2} f(x)dx$$
If f(x) is odd, then what can we say about $$\int_{-2}^{2} f(x)dx$$ 
I guessed they both are zero? For the first one if its even wouldn't this be the same as $$\int_{a}^{a} f(x)dx = 0$$
Now if its odd f(-x) = -f(x). Would FTOC make this zero as well? 

Not exactly:
$$\begin{cases}\displaystyle\int_{-a}^a f(x)\,\mathrm d\mkern1mu x=2\int_{0}^a f(x)\,\mathrm d\mkern1mu x &\text{if }\;f\;\text{ is even,}\\
\displaystyle\int_{-a}^a f(x)\,\mathrm d\mkern1mu x=0&\text{if }\;f\;\text{ is odd.}\end{cases}$$
To see it, make the substitution $\;t=-x$, $\;\mathrm d\mkern1mu x=-\mathrm d\mkern1mu t$:
$$\int_{-a}^0 f(x)\,\mathrm d\mkern1mu x=-\int_{a}^0 f(-t)\,\mathrm d\mkern1mu t=\int_{0}^a f(-t)\,\mathrm d\mkern1mu t=\begin{cases}\displaystyle\int_{0}^a f(-t)\,\mathrm d\mkern1mu t&(f\;\text{even}),\\\displaystyle-\int_{0}^a f(-t)\,\mathrm d\mkern1mu t&(f\;\text{odd}),\end{cases}$$
then use Chasles relation.
