Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 882: 23

$e-2$

Consider: $I= \int_{0}^{1} \int_{0}^{y^2} 3 y^3 e^{xy} dx dy$ or, $= \int_{0}^{1} [3y^2 e^{xy}]_{0}^{y^2} dy $ or, $= \int_0^{1} 3y^2(e^{y^3}-1) dy $ Set $a=y^3 \implies da= 3y^2dy$ So, $I=\int_0^1 e^a-1 da =[e^a-a]_0^1=e^1-1-1-0=e-2$