Answer
$36$
Work Step by Step
Let us suppose that $I=\int_{0}^6 \int_{y/2}^{3} (x+y) \ dx \ dy$
Next, we will solve the integral as follows:
$I=\int_{0}^6 [\int_{y/2}^{3} (x+y) \ dx] \ dy$
or, $=\int_{0}^6 [\dfrac{x^2}{2}+xy]_{y/2}^{3} \ dy$
or, $=\int_{0}^6 (\dfrac{9}{2}+3y-\dfrac{y^2}{8}-\dfrac{y^2}{2}) \ dy$
or, $= (\dfrac{9y}{2}+\dfrac{3y^2}{2}-\dfrac{5y^3}{24}]_{0}^6$
or, $=\dfrac{9(6)}{2}+\dfrac{3(36)}{2}-\dfrac{1080}{24}$
or, $=36$
