## Thomas' Calculus 13th Edition

$\dfrac{-1}{10}$
After integrating the inner $dv$ integral, we get: $\iint_{R} f(u,v) dA= \int_{0}^{1} [\dfrac{v^2}{2}-vu^{1/2}]_0^{1-u} du$ or, $= \int_{0}^{1 } [1/2) (1-u)^2-(1-u) \sqrt u ] du$ or, $= \int_{0}^{1 } \dfrac{1-4x^3-3x+6x^2}{3} dx$ or, $= \dfrac{1}{2}+\dfrac{1}{6}-\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{2}{5}$ Thus, we have $I=\dfrac{-1}{10}$