Answer
$$\dfrac{1}{5}$$
Work Step by Step
$$I=\int_{0}^{1} \int_{\sqrt y}^{2-\sqrt y} xy \ dx \ dy \\=[\dfrac{x^2}{2}] \times \int_{0}^{1} [x^2]_{\sqrt y}^{2-\sqrt y}\ dy \\=\dfrac{1}{2} \int_{0}^{1} (2-\sqrt y)^2 \times y-(\sqrt y)^2 y \ dy \\=\dfrac{1}{2} \int_{0}^{1} (4-4\sqrt y+y) \times y- y^2 \ dy \\=\int_0^1 2y-2y^{3/2} \ dy \\=[y^2-\dfrac{4y^{5/2}}{5}]_0^1 \\=\dfrac{1}{5}$$
