TITLE: Definite integral of an odd function is 0 (symmetric interval)
QUESTION [16 upvotes]: For an odd function, I know that f(x) = - f(x).
I'm trying to show that $\int^{a}_{-a} f(x) dx$ = 0. 
I've seen the proof where it splits the integral up into:
$$\int^{a}_{0} f(x) dx + \int^{0}_{-a} f(x) dx $$ However, I still don't understand how to evaluate the second part, which is the 'gold' of the proof.
For example in https://proofwiki.org/wiki/Definite_Integral_of_Odd_Function, why do we define a function from $x \mapsto -x$? (hence leading onto du = - dx and so on). I've been to different sites/videos and they all show me 'how', but not 'why' certain steps are carried out.
Additionally, why can't I evaluate the original integral right away? 
$$\int^{a}_{-a} f(x) dx = [F(x)]^{a}_{-a} = F(a) - F(-a) = F(a) - F(a) = 0 $$? Is it because we're limited by the need to prove that the integral of that (odd) function is even?

REPLY [21 votes]: Because $f$ is odd, $f(-x) = -f(x)$.  Thus, we can write
$$
\int_{x=-a}^0 f(x) \, dx = \int_{x=-a}^0 -f(-x) \, dx
$$
Let $u = -x, du = -dx$, then
$$
\int_{x=-a}^0 -f(-x) \, dx = \int_{u=a}^0 f(u) \, du
$$
Reversing the limits of integration inverts the result, so
$$
\int_{u=a}^0 f(u) \, du = -\int_{u=0}^a f(u) \, du
$$
as desired.  (The $u$ is just a dummy variable; it can be replaced by $x$ at this point, despite our originally having assigned it as $u = -x$.)