The rectangle [tex]R[/tex] is the set
[tex]R = \{(x,y) \mid 0 \le x \le 2 \text{ and } -1 \le y \le 1\}[/tex]
The average value of [tex]f(x,y)[/tex] over [tex]R[/tex] is the ratio of the integral of [tex]f[/tex] over [tex]R[/tex] to the area of [tex]R[/tex].
[tex]f_{\rm ave} = \dfrac{\displaystyle \iint_R f(x,y) \, dA}{\displaystyle \iint_R dA}[/tex]
The rectangle is 2-by-2, so its area is
[tex]\displaystyle \iint_R dA = 2\times2 = 4[/tex]
Integrate [tex]f[/tex].
[tex]\displaystyle \iint_R xy^2 \, dA = \int_{-1}^1 \int_0^2 xy^2 \, dx \, dy \\\\ ~~~~~~~~ = 2 \int_{-1}^1 y^2 \, dy \\\\ ~~~~~~~~ = 4 \int_0^1 y^2 \, dy = \frac43[/tex]
Then the average value of [tex]f[/tex] is
[tex]f_{\rm ave} = \dfrac{\frac43}4 = \boxed{\dfrac13}[/tex]
