integrals of even and odd functions

Theorem.  Let the real function $f$ beRiemann-integrable (http://planetmath.org/RiemannIntegrable) on $[-a,a]$.  If $f$ is an

• even function, then ${\displaystyle {\int }_{-a}^a}f(x)𝑑x=\mathrm{\hspace{0.33em}2}{\displaystyle {\int }_0^a}f(x)𝑑x$,

• odd function, then ${\displaystyle {\int }_{-a}^a}f(x)𝑑x=\mathrm{\hspace{0.33em}0}.$

Of course, both cases concern the zero map which is both.

Proof. Since the definite integral is additive with respect to the interval of integration, one has

$$I:={\int }_{-a}^{a}f(x)𝑑x={\int }_{-a}^{0}f(t)𝑑t+{\int }_0^{a}f(x)𝑑x.$$

Making in the first addend the substitution  $t=-x,dt=-dx$  and swapping the limits of integration one gets

$$I={\int }_a^{0}f(-x)(-dx)+{\int }_0^{a}f(x)𝑑x={\int }_0^{a}f(-x)𝑑x+{\int }_0^{a}f(x)𝑑x.$$

Using then the definitions of even (http://planetmath.org/EvenoddFunction) ($+$) and odd (http://planetmath.org/EvenoddFunction) ($-$) function yields

$$I={\int }_0^a(\pm f(x))𝑑x+{\int }_0^{a}f(x)𝑑x=\pm {\int }_0^{a}f(x)𝑑x+{\int }_0^{a}f(x)𝑑x,$$

which settles the equations of the theorem.