Chapter 14 - Multiple Integrals - 14.1 Double Integrals - Exercises Set 14.1 - Page 1007: 7

$\int_{-1}^{0} \int_{2}^{5} \ d x \ d y=3$,

Given $$\int_{-1}^{0} \int_{2}^{5} \ d x \ d y$$ So, we get \begin{aligned}I &=\int_{0}^{1} \int_{0}^{2} \ d x \ d y\\ &=\int_{2}^{5} \ d x \int_{-1}^{0} \ d y \\ &=[ x]_{2}^{5} \ \ \left[y\right]_{-1}^{0} \\ &=(5-2)(0+1)\\ &=3 \end{aligned}