Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 882: 26

$\dfrac{1}{6}$

After integrating the inner $dy$ integral, we get: $I= \int_{0}^{1} [x^2y+\dfrac{y^3}{3}]_0^{1-x} dx$ or, $= \int_{0}^{1 } x^2(1-x) y+\dfrac{(1-x)^3}{3} dx$ or, $= \int_{0}^{1 } \dfrac{1-4x^3-3x+6x^2}{3} dx$ or, $= [ \dfrac{x-x^4-3(x^2/2)+2x^3}{3}]_0^1 dx$ Thus, we have $I=\dfrac{1}{6}$