Even and Odd Functions over Symmetric Limits

When the limits of integration are symmetric about $0$ — that is, of the form $-a$ to $a$ — the symmetry of the integrand can be exploited to shorten or immediately settle the integral.

Even and odd functions. A function is even if $f(-x) = f(x)$ for all $x$ (its graph is symmetric about the $y$-axis; e.g. $x^{2}$, $x^{4}$, any constant, $x^2+1$). It is odd if $f(-x) = -f(x)$ (its graph has half-turn symmetry about the origin; e.g. $x$, $x^{3}$, $x^{5}$).

The property (P7).

For an even integrand the two halves of the region (left and right of the $y$-axis) are mirror images with equal area, so the whole integral is twice the right half. For an odd integrand the left half is the negative of the right half, so they cancel and the integral is exactly $0$.

Method — using symmetry over $[-a, a]$:

1. Test the integrand: replace $x$ by $-x$ and simplify. If the result equals $f(x)$ it is even; if it equals $-f(x)$ it is odd.
2. If odd, write down $0$ immediately.
3. If even, evaluate $2\displaystyle\int_{0}^{a} f(x)\,dx$ (often easier, since the lower limit is $0$).
4. If the function is neither even nor odd, the property does not apply — evaluate the integral $\displaystyle\int_{-a}^{a}$ directly.

Illustration. $\displaystyle\int_{-2}^{2} x^{3}\,dx$: here $f(x)=x^{3}$ and $f(-x) = (-x)^{3} = -x^{3} = -f(x)$, so $f$ is odd and the integral is $0$ — no antiderivative needed. (Directly: $\big[x^{4}/4\big]_{-2}^{2} = \tfrac{16}{4} - \tfrac{16}{4} = 0$, confirming it.)