# Solución

Primero encuentra el área $A (D)$ donde la región $D$ está dada por la figura. Tenemos

$$\begin{aligned} A(D) &= \iint_D1dA = \int_{y=0}^{y=1}\int_{x=y}^{x=\sqrt{y}}1dxdy = \int_{y=0}^{y=1}\bigg[x\bigg|_{x=y}^{x=\sqrt{y}}\bigg]dy\\ &= \int_{y=0}^{y=1}(\sqrt{y} - y)dy = \frac23y^{3/2} - \frac{y^2}{2}\bigg|_0^1 = \frac16 \end{aligned}$$

Entonces el valor promedio de la función dada sobre esta región es

$$\begin{aligned} f_{prom} &= \frac{1}{A(D)}\iint_D f(x, y)dA = \frac{1}{A(D)}\int_{y=0}^{y=1}\int_{x=y}^{x=\sqrt{y}}7xy^2dxdy\\ &= \frac{1}{1/6}\int_{y=0}^{y=1}\bigg[\frac72x^2y^2\bigg|_{x=y}^{x=\sqrt{y}}\bigg]dy = 6\int_{y=0}^{y=1}\bigg[\frac72y^2\big(y-y^2\big)\bigg]dy\\ &= 6\int_{y=0}^{y=1}\bigg[\frac72\big(y^3-y^4\big)\bigg]dy = \frac{42}{2}\bigg(\frac{y^4}{4} - \frac{y^5}{5}\bigg)\bigg|_0^1 = \frac{42}{40} = \frac{21}{20} \end{aligned}$$