Definite Integral of Odd Function

Theorem
Let $f$ be an odd function with a primitive on the closed interval $\left[{-a \,.\,.\, a}\right]$, where $a > 0$.

Then:
 * $\displaystyle \int_{-a}^a f \left({x}\right) \ \mathrm d x = 0$

Proof
Let $F$ be a primitive for $f$ on the interval $\left[{-a \,.\,.\, a}\right]$.

Then, by Sum of Integrals on Adjacent Intervals, we have:

Therefore, it suffices to prove that:
 * $\displaystyle \int_{-a}^0 f \left({x}\right) \ \mathrm d x = - \int_0^a f \left({x}\right) \ \mathrm d x$

To this end, let $\phi: \R \to \R$ be defined by $x \mapsto -x$.

From Derivative of Identity Function and Derivative of Constant Multiple, for all $x \in \R$, we have $\phi' \left({x}\right) = -1$.

Then, by means of Integration by Substitution, we compute:

This concludes the proof.

Also see

 * Definite Integral of Even Function