# What does this definite integral theorem mean?

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

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

• 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 $\int_{-a}^a f(x)dx=\int_{0}^a(f(x)+f(-x))dx$. – Yourong 'DZR' Zang Jan 16 at 4:22
• Ohh okay, thank you so much! – Keira Evangeline Jan 16 at 4:27
• 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 – QED Jan 16 at 5:39
• 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. – CyCeez Jan 16 at 9:32
• Oh, okay. Thank you so much for the advice! – 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:

$$\hskip 4.5cm$$

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}

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