Solución

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

A(D)=∬D1dA=∫y=0y=1∫x=yx=y1dxdy=∫y=0y=1[x∣x=yx=y]dy=∫y=0y=1(y−y)dy=23y3/2−y22∣01=16\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

fprom=1A(D)∬Df(x,y)dA=1A(D)∫y=0y=1∫x=yx=y7xy2dxdy=11/6∫y=0y=1[72x2y2∣x=yx=y]dy=6∫y=0y=1[72y2(y−y2)]dy=6∫y=0y=1[72(y3−y4)]dy=422(y44−y55)∣01=4240=2120\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}