Answer
$\dfrac{37}{6}$
Work Step by Step
Consider $Area; A=\int_{1}^{4} \int_{2-y}^{\sqrt y} \ dx \ dy$
or, $=\int_{1}^{4} (\sqrt y-2+y) \ dy $
or, $=\int_{1}^{4} \sqrt y dy - \int_{1}^{4} 2 \ dy + \int_{1}^{4} y \ dy $
or, $=[ \dfrac{2y^{3/2}}{3}]_{1}^{4} - [ 2y]_{1}^{4}+[\dfrac{y^2}{2}]_{1}^4 $
or, $=\dfrac{14}{3}-6+\dfrac{15}{2}$
or, $=\dfrac{37}{6}$
