Definite Integral of Even Function

Theorem
Let $f$ be an even function with a primitive on the closed interval $\left[{-a..a}\right]$.

Then
 * $\displaystyle \int_{-a}^a f(x) \ \mathrm d x = 2 \int_0^a f \left({x}\right) \ \mathrm d x$

Proof
$\int_{-a}^{a}f(x)\mathrm{d}{x} =

\int_{-a}^{0}f(x)\mathrm{d}{x} + \int_{0}^{a}f(x)\mathrm{d}{x}$

$ = -1\int_{0}^{-a}f(x)\mathrm{d}{x} + \int_{0}^{a}f(x)\mathrm{d}{x}$

Define $u ≡ -x$. Then $du = -dx$. Also,

$-x = 0 \iff u = 0$.

$x = -a \iff u = a$

$f(x) = f(-u)$

$\int_{-a}^{a}f(x)\mathrm{d}{x} = -1\int_{0}^{a}f(-u)\mathrm{(-d}{u)} + \int_{0}^{a}f(x)\mathrm{d}{x}$

$= \int_{0}^{a}f(-u)\mathrm{d}{u} + \int_{0}^{a}f(x)\mathrm{d}{x}$

Because we defined $f$ to be even, $f(-u)=f(u)$.

$= \int_{0}^{a}f(u)\mathrm{d}{u} + \int_{0}^{a}f(x)\mathrm{d}{x}$

Because $u$ is just a dummy variable, we can replace all instances of it with $x$.

$= \int_{0}^{a}f(x)\mathrm{d}{x} + \int_{0}^{a}f(x)\mathrm{d}{x}$

$ = 2\int_{0}^{a}f(x)\mathrm{d}{x}$