Definite Integral of Odd Function

Theorem
Let $f$ be an odd function with a primitive on the open interval $\closedint {-a} a$, where $a > 0$.

Then:
 * $\displaystyle \int_{-a}^a \map f x \rd x = 0$

Proof
Let $F$ be a primitive for $f$ on the interval $\closedint {-a} a$.

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

Therefore, it suffices to prove that:
 * $\displaystyle \int_{-a}^0 \map f x \rd x = -\int_0^a \map f x \rd x$

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 $\map {\phi'} x = -1$.

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

This concludes the proof.

Also see

 * Definite Integral of Even Function