0

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$$\int_{-a}^{a}f(x)dx=0$$

Using Riemann Sum for proving

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!

Martin Sleziak
  • 53,687
jaclynx
  • 121
  • 1
    It is not to show that f is odd, this is a condition. – Peter Oct 19 '16 at 09:55
  • Yes, @Yongyong is correct. You are told a priori that $f$ is odd, therefore use a Riemann sum approach to show that the integral over a symmetric range identically vanishes. –  Oct 19 '16 at 09:56
  • Also, if you are sure that $f$ is indeed integrable on $[-a,a]$ then you should show (or, I suppose explain) that $f$ is separably integrable on $[-a,0]$ and $[0,a]$. So my advice would be partition on $[-a,0]$ and explain how to transform into Riemann Sums on $[0,a]$. –  Oct 19 '16 at 09:59

1 Answers1

4

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

$$ \begin{align} \int_{-a}^{0}{f(x)}\,dx &= \lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(-x_{i}^{*})\Delta x}} \\ &= \lim_{n \to \infty}{\sum\limits_{i=1}^{n}{-f(x_{i}^{*})\Delta x}} \\ &= -\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}} \\ \end{align} $$

$$ \begin{align} \therefore \int_{-a}^{a}f(x)\,dx&=\int_{-a}^{0}{f(x)}\,dx+\int_{0}^{a}{f(x)}\,dx \\ &= -\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}}+\lim_{n \to \infty}{\sum\limits_{i=1}^{n}{f(x_{i}^{*})\Delta x}} \\ &= 0 \\ \end{align} $$

Skeleton Bow
  • 2,101