The FunctionAverage command returns the average value of a function over a region in one-, two-, or three-dimensional space.

The opts argument can contain any of the following equations that set output options.

coordinates = cartesian[x,y], polar[r,theta] (2-D), cartesian[x,y,z], cylindrical[r,theta,z], or spherical[r,phi,theta]

Determines the coordinate system being used. The first variable of polar, cylindrical, and spherical is assumed to be the radial component. The default is cartesian.

output = value or integral

This option controls the return value of the function.

* output = value specifies that the average value of the function is returned. The default is output = value.

* output = integral specifies that the inert form of the function average is returned.

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\right)\:$

$\mathrm{FunctionAverage}\left(x^3\,x=3..4\right)$

$\frac{175}{4}$

$\mathrm{FunctionAverage}\left(x^2+y^2\,y=0..\mathrm{sqrt}\left(1-x^2\right)\,x=0..1\right)$

$\frac{1}{2}$

$\mathrm{FunctionAverage}\left(r^2\,r=0..1\,t=0..\frac{\mathrm{\pi }}{2}\,\mathrm{coordinates}=\mathrm{polar}\left[r,t\right]\right)$

$\frac{1}{2}$

$\mathrm{FunctionAverage}\left(x^2+y^2+z\,z=-2..4+y^2\,y=x-1..x+6\,x=2..4\,\mathrm{output}=\mathrm{integral}\right)$

$\frac{{\int }_2^4\left({\int }_{x-1}^{x+6}\left({\int }_{\mathrm{-2}}^{y^2+4}\left(x^2+y^2+z\right)\ⅆz\right)\ⅆy\right)\ⅆx}{{\int }_2^4\left({\int }_{x-1}^{x+6}\left({\int }_{\mathrm{-2}}^{y^2+4}1\ⅆz\right)\ⅆy\right)\ⅆx}$

$\mathrm{FunctionAverage}\left(x^2+y^2\,y=x^2+4..x^3\,x=a..b\right)$

$\frac{-\frac{64}{3}b+\frac{64}{3}a-\frac{1}{30}{a}^{10}+\frac{1}{30}{b}^{10}+\frac{1}{21}{a}^7-\frac{1}{21}{b}^7-\frac{1}{6}{a}^6+\frac{1}{6}{b}^6+a^5-b^5+\frac{20}{3}{a}^3-\frac{20}{3}{b}^3}{-\frac{1}{4}{a}^4+\frac{1}{4}{b}^4+\frac{1}{3}{a}^3-\frac{1}{3}{b}^3-4b+4a}$

$\mathrm{FunctionAverage}\left(x^2+y^2\,y=x^2+4..x^3\,x=0..4\right)$

$\frac{218656}{175}$

$\mathrm{FunctionAverage}\left(x^2+y^2\,y=x^2+4..x^3\,x=3..5\right)$

$\frac{16739158}{5005}$