Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 759: 6

$0$

Re-arrange the given integral as follows: $\int_{0}^{3}[\dfrac{x^2y^2}{2}-xy^2]_{-2}^{0}]dx=\int_{0}^{3}[4x-2x^2] dx$ This implies that $[2x^2-\dfrac{2x^3}{3}]_{0}^{3}= 2(3)^2-\dfrac{2(3)^3}{3}$ Thus, we have, $18-18=0$