Answer
$$\dfrac{37}{6}$$
Work Step by Step
We calculate the integral as follows:
$$ Area=\int_{1}^{4} \int_{2-y}^{\sqrt y} \ dx \ dy \\=\int_{1}^{4} (y^{1/2}-2+y) \ dy \\=\int_{1}^{4} \sqrt y dy - \int_{1}^{4} 2 \ dy + \int_{1}^{4} y \ dy \\=[ \dfrac{2}{3} y^{3/2}]_{1}^{4} - 2[y]_{1}^{4}+[\dfrac{1}{2} \times y^2]_{1}^4 \\=\dfrac{14}{3}-6+\dfrac{15}{2} \\=\dfrac{37}{6}$$
