As I was solving a homework problem about definite integrals, I came across a theorem to help me solve the problem. Although I got the right answer to the problem, I do not really understand the theorem itself. I would be grateful if you provide me with a short explanation about what the theorem really means and when I could use it. Thank you!

Here is the theorem: picture of theorem

Theorem from image stated: If $f(x)$ is an odd function and continuous on $[-a,a]$, then

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

  • $\begingroup$ If you approach this integral as a Riemann sum then you will see that terms $f(x)$ and $f(-x)$ cancel and therefore the limit (integral) will be $0$. Or else you could write $\int_{-a}^a f(x)dx=\int_{0}^a(f(x)+f(-x))dx$. $\endgroup$ – Yourong 'DZR' Zang Jan 16 at 4:22
  • $\begingroup$ Ohh okay, thank you so much! $\endgroup$ – Keira Evangeline Jan 16 at 4:27
  • $\begingroup$ you could as use the Fundamental Theorem since the antiderivative of a odd function is even (you can prove this using the chain rule), so F(a)-F(-a)=0 where F is the antiderivative of f $\endgroup$ – QED Jan 16 at 5:39
  • $\begingroup$ Welcome! Just FYI, it's always good to type out the problem statement, so that when people use the search feature, they can find problems like this. $\endgroup$ – CyCeez Jan 16 at 9:32
  • $\begingroup$ Oh, okay. Thank you so much for the advice! $\endgroup$ – Keira Evangeline Jan 16 at 17:36

I'm unsure what level of calculus you're at, but first, we can give some intuitive reasoning. If $f$ is an odd function, then $f(-x)=-f(x)$ for all $x$ in the domain of $f$ or equivalently, $f(x) = -f(-x)$. This simply means when $x < 0$, the function $f$ evaluates to the negative of $f$ values when $x > 0$. So, then the graph of an odd function will look something like the graph below:

$\hskip 4.5cm$ enter image description here

So, the graph of $f$ has a "negative symmetry," which I mean that the graph of $f$ for positive $x$-values are reflected across the $y$-axis and then negated (or multiplied by $-1$). Since, the definite integral is the signed area between the graph and the $x$-axis, then you can see that the areas on both sides of the $y$-axis are the same except that the definite integral from $-a$ to $0$ should be negative, and so the values of the definite integrals on both sides of the $y$-axis cancel out.

More analytically speaking, we can use $f(x) = -f(-x)$ to write

\begin{align} \int_{-a}^a f(x) \, dx &= \int_{-a}^0 f(x) \, dx + \int_0^a f(x) \, dx \\ &= -\int_{-a}^0 f(-x) \, dx + \int_0^a f(x) \, dx. \end{align}

Substituting $u = -x$ (and hence $-du = dx$) in the first integral on the RHS,

\begin{align} \int_{-a}^a f(x) \, dx &= \int_a^0 f(u) \, du + \int_0^a f(x) \, dx \\ &= -\int_0^a f(u) \, du + \int_0^a f(x) \, dx \\ &= 0. \end{align}

  • $\begingroup$ Thank you so much for your help! It's much clearer now. $\endgroup$ – Keira Evangeline Jan 16 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.