average value of function

The set of the values of a real function $f$ defined on an interval  $[a,b]$  is usually uncountable, and therefore for being able to speak of an average value of $f$ in the sense of the average value

$A.V.={\displaystyle \frac{a_1+a_2+\mathrm{\dots }+a_n}{n}}={\displaystyle \frac{{\sum }_{j=1}^n{a}_j}{{\sum }_{j=1}^{n}1}}$(1)

of a finite list  $a_1,a_2,\mathrm{\dots },a_n$  of numbers, one has to replace the sums with integrals.  Thus one could define

$$A.V.(f):=\frac{{\int }_a^{b}f(x)𝑑x}{{\int }_a^{b}1𝑑x},$$

i.e.

$A.V.(f):={\displaystyle \frac{1}{b-a}}{\displaystyle {\int }_a^b}f(x)𝑑x.$(2)

For example, the average value of $x^2$ on the interval  $[0,\mathrm{\hspace{0.17em}1}]$  is $\frac{1}{3}$ and the average value of $\sin x$on the interval  $[0,\pi ]$  is $\frac{2}{\pi }$.

The definition (2) may be extended to complex functions $f$ on an arc $\gamma$ of a rectifiable curve via the contour integral

$A.V.(f):={\displaystyle \frac{1}{l(\gamma )}}{\displaystyle {\int }_{\gamma }}f(z)𝑑z$(3)

where $l(\gamma )$ is the length (http://planetmath.org/ArcLength) of the arc.  If especially $\gamma$ is a closed curve in a simply connected domain where $f$ is analytic, then the average value of $f$ on $\gamma$ is always 0, as the Cauchy integral theorem implies.