What does this definite integral theorem mean?

Question

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

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

Answer 1

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:

$\hskip 4.5cm$

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}