Show $\int_{-a}^{a} [f(x)-f(-x)] dx=0$ for any continuous function f(x) Show that for any continuous function $f(x)$ on $[-a, a]$, where $a>0$,
$$\int_{-a}^{a} [f(x)-f(-x)] dx=0$$

I've tried to do something like:
Let the primitive function of $f(x)$ be $F(x)$.
$\int_{-a}^{a} [f(x)-f(-x)] dx$
$=[F(x)]_{-a}^{a}-[F(-x)]_{-a}^{a}$
$=F(-a)-F(a)-F(a)+F(-a)$
$=2F(-a)-2F(a)$
but I don't know how to continue.
 A: Hint:
$$I=\int_a^bf(x) \ dx=\int_a^bf(a+b-x)\ dx$$
$$I+I=\int_a^b\{f(x) +f(a+b-x)\}\ dx$$
A: Hint
If $g$ is an odd function then
$$\int_{-b}^b g(t)dt=0$$
A: If $g$ is continuous, if $\phi$ and $\psi$ are differentiable, and if
$$
G(x) = \int_{\psi(x)}^{\phi(x)} g(t)\, dt,
$$
the fundamental theorem of calculus and the chain rule give
$$
G'(x) = g\bigl(\phi(x)\bigr)\, \phi'(x) - g\bigl(\psi(x)\bigr)\, \psi'(x).
\tag{1}
$$
Define
$$
F(x) = \int_{-x}^{x} \bigl[f(t) - f(-t)\bigr]\, dt.
$$
The integrand $g(x) = f(x) - f(-x)$ is continuous (since $f$ is continuous), so (1) becomes
$$
F'(x) = \bigl[f(x) - f(-x)\bigr] - \bigl[f(-x) - f(x)\bigr](-1)
  = 0
$$
for all $x$ in $[-a, a]$. By the identity theorem, $F(x) = F(0) = 0$ for all $x$ in $[-a, a]$.
A: we have:
$$
\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx-\int_{-a}^af(-x)dx
$$
and, substituting $-x=t \rightarrow dx=-dt$ in the second integral:
$$
\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx+\int_{a}^{-a}f(t)dt
$$
that is (since the variable in the integral is ''free''):
$$
\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx+\int_{a}^{-a}f(x)dx
=0$$
A: Hint: Substitute $x=-t$,
$$
\begin{align}
\underbrace{\int_{-a}^a\left[f(x)-f(-x)\right]\,\mathrm{d}x}_{I}
&=\int_a^{-a}\left[f(-t)-f(t)\right]\,\mathrm{d}(-t)\\
&=\underbrace{\int_{-a}^a\left[f(-t)-f(t)\right]\,\mathrm{d}t}_{-I}
\end{align}
$$
