Definite Integral of Even Function
Theorem
Let $f$ be an even function with a primitive on the closedinterval $\closedint {-a} a$, where $a > 0$.
Then:
- $\ds \int_{-a}^a \map f x \rd x = 2 \int_0^a \map f x \rd x$
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:
- $\ds \int_{\mathop \to -a}^{\mathop \to a} \map f x \rd x = 2 \int_0^{\mathop \to a} \map f x \rd x$
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:
\(\ds \int_{-a}^a \map f x \rd x\) | \(=\) | \(\ds \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x\) |
Therefore, it suffices to prove that:
- $\ds \int_{-a}^0 \map f x \rd x = \int_0^a \map f x \rd x$
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:
\(\ds \int_{\map \phi a}^{\map \phi 0} \map f x \rd x\) | \(=\) | \(\ds \int_a^0 \map f {-u} \paren {-1} \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^a \map f {-u} \rd u\) | Definition of Definite Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^a \map f {-x} \rd x\) | renaming integration variable | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^a \map f x \rd x\) | $f$ is an even function |
This concludes the proof.
$\blacksquare$
Also see
Sources
- 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.): $\S 4.5$