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

Symmetry Property of Definite Integrals

Suppose you must find $\int_{-2}^{2} x^3\,dx$. You could integrate directly — or you could notice the graph of $x^3$ is anti-symmetric about the origin, so every positive bit of area on the right is cancelled by an equal negative bit on the left, and the answer is simply $0$. When the interval is symmetric about zero, the symmetry of the function does the work for you.

Even and odd functions

• An even function satisfies $f(-x) = f(x)$ (e.g. $x^2$, $\cos x$, $|x|$). Its graph is a mirror image across the $y$-axis, so the area on $[-a,0]$ equals the area on $[0,a]$.
• An odd function satisfies $f(-x) = -f(x)$ (e.g. $x^3$, $\sin x$). Its left half is the negative mirror of its right half, so the two areas cancel.

Why it works

Split at zero and substitute $u = -x$ in the left piece:

If $f$ is even, $f(-u) = f(u)$ and the two integrals add to $2\int_0^a f$. If $f$ is odd, $f(-u) = -f(u)$ and they cancel to $0$.

Using it

$\int_{-3}^{3} x^4\,dx = 2\int_0^3 x^4\,dx = 2\left[\tfrac{x^5}{5}\right]_0^3 = \tfrac{486}{5}, \qquad \int_{-\pi}^{\pi} \sin x\,dx = 0.$ …