Let $f$ be a continuous function on $[-a,a]$, where $a>0$. Use Riemann sums to prove that if $f$ is an odd function, then$$\int_{-a}^{a}f(x)dx=0$$
Very unsure about how to use Riemann sum to prove that f is an odd function. Any solutions/hints on how to get started on this question will be greatly appreciated!
If $f(x)$ is odd, then $f(-x)=-f(x), \; \forall x \in \mathbb R$.
$$\int_{-a}^{a}f(x)\,dx=\int_{-a}^{0}{f(x)}\,dx+\int_{0}^{a}{f(x)}\,dx$$ Now $$\int_{0}^{a}{f(x)}\,dx=\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}}$$ But
$$ \begin{align} \int_{-a}^{0}{f(x)}\,dx &= \lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(-x_{i}^{*})\Delta x}} \\ &= \lim_{n \to \infty}{\sum\limits_{i=1}^{n}{-f(x_{i}^{*})\Delta x}} \\ &= -\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}} \\ \end{align} $$
$$ \begin{align} \therefore \int_{-a}^{a}f(x)\,dx&=\int_{-a}^{0}{f(x)}\,dx+\int_{0}^{a}{f(x)}\,dx \\ &= -\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}}+\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}} \\ &= 0 \\ \end{align} $$