Definite Integral of Even Function

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Theorem

Let $f$ be an even function with a primitive on the closedinterval $\closedint {-a} a$, where $a > 0$.


Then:

$\displaystyle \int_{-a}^a \map f x \rd x = 2 \int_0^a \map f x \rd x$


Corollary

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


Then the improper integral of $f$ on $\openint {-a} a$ is:

$\displaystyle \int_{\mathop \to -a}^{\mathop \to a} \map f x \rd x = 2 \int_0^{\mathop \to a} \map f x \rd x$


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:

\(\displaystyle \int_{-a}^a \map f x \rd x\) \(=\) \(\displaystyle \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x\)

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:

\(\displaystyle \int_{\map \phi a}^{\map \phi 0} \map f x \rd x\) \(=\) \(\displaystyle \int_a^0 \map f {-u} \paren {-1} \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^a \map f {-u} \rd u\) Definition of Definite Integral
\(\displaystyle \) \(=\) \(\displaystyle \int_0^a \map f {-x} \rd x\) renaming integration variable
\(\displaystyle \) \(=\) \(\displaystyle \int_0^a \map f x \rd x\) $f$ is an even function

This concludes the proof.

$\blacksquare$


Also see


Sources