Theorem. Let the real function $f$ be Riemann-integrable on $[-a,\,a]$ . If $f$ is an

• even function, then $\displaystyle \int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$ ,
• odd function, then $\displaystyle \int_{-a}^a f(x)\,dx = 0.$

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 ($+$ ) and odd ($-$ ) 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.