Answer
$$\dfrac{5}{6}$$
Work Step by Step
$$\iint_{R} dA= \int_{1}^{2} \int_{y}^{y^2} dx dy\\= \int_1^{2} [x]_{y}^{y^2} dy \\= \int_1^{2} [y^2-y] dy \\=[\dfrac{y^3}{3}-\dfrac{y^2}{2}]_1^2 \\=[\dfrac{8}{3} -2]-[\dfrac{1}{3}+\dfrac{1}{2}] \\=\dfrac{5}{6}$$
