Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 759: 15

$14$

Re-arrange the given integral as follows: $\int_{0}^{1}\int_{0}^{2}(6y^2-2x) dy dx=\int_{0}^{2} (6y^2x)_0^1-x^2|_0^1 dy$ This implies that $\int_{0}^{2} (6y^2x)_0^1-x^2|_0^1 dy=\int_{0}^{2} (6y^2-1) dy$ Hence, $[\dfrac{6y^3}{3}-y]_0^2=[\dfrac{6(2)^3}{3}-\dfrac{6(0)^3}{3}]-(2-0)=14$