Answer
The average value of $f$ is $\frac{5}{6}$.
Work Step by Step
The region $R$ is a rectangle with vertices $(-1,0)$, $(-1,5)$, $(1,5)$, and $(1,0)$ or it can be written as the set of points $R=\{(x,y)|-1\leq x\leq 1, 0\leq y\leq 5\}$.
The area of $R$ is $A=length\times width=\Delta x\cdot\Delta y=(1-(-1))\cdot (5-0)=10$.
Find the average value of $f$ over $R$:
$\overline{f}=\frac{1}{A}\iint_R f(x,y)dA$
$=\frac{1}{10}\int_{-1}^1\int_0^5x^2ydydx$
$=\frac{1}{10}\int_{-1}^1\frac{x^2y^2}{2}]_0^5dx$
$=\frac{1}{10}\int_{-1}^1\frac{25x^2}{2}dx$
$=\frac{1}{10}[\frac{25x^3}{6}]_{-1}^1$
$=\frac{1}{10}(\frac{25}{6}-(-\frac{25}{6}))$
$=\frac{5}{6}$
