Thomas' Calculus 13th Edition

$\dfrac{5}{6}$
Consider: $I= \int_{1}^{2} \int_{y}^{y^2} dx dy$ or, $= \int_1^{2} [x]_{y}^{y^2} dy$ or, $= \int_1^{2} [y^2-y] dy$ or, $=[\dfrac{y^3}{3}-\dfrac{y^2}{2}]_1^2$ So, $I=\dfrac{8}{3} -2-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}$