Answer
The average value of $f$ over $R$ is $\frac{(4+e)^{5/2}-e^{5/2}-32}{15}$.
Work Step by Step
The area of $R$ is $A=length\cdot width=4\cdot 1=4$.
Find the double integral $\iint_R f(x,y)dA$:
$\int_{0}^4\int_0^1e^y\sqrt{x+e^y}dydx=\int_0^4\frac{2(x+e^y)^{3/2}}{3}]_0^1dx=\int_0^4\frac{2(x+e)^{3/2}}{3}-\frac{2x^{3/2}}{3}dx=\frac{4(x+e)^{5/2}}{15}-\frac{4x^{5/2}}{15}]_0^4=\frac{4(4+e)^{5/2}}{15}-\frac{4e^{5/2}}{15}-\frac{128}{15}=\frac{4((4+e)^{5/2}-e^{5/2}-32)}{15}$
Find the average value of $f$ over $R$:
$\bar{f}=\frac{1}{A}\iint_R f(x,y)dA=\frac{1}{4}\cdot \frac{4((4+e)^{5/2}-e^{5/2}-32)}{15}=\frac{(4+e)^{5/2}-e^{5/2}-32}{15}$
