Chapter 14 - Multiple Integrals - 14.2 Double Integrals Over Nonrectangular Regions - Exercises Set 14.2 - Page 1016: 33

False.

Using the rules of integrals, we obtain: $\int_{1}^{2} \int_{y / 2}^{1} f(x, y) d x d y+\int_{0}^{1} \int_{y / 2}^{\sqrt{y}} f(x, y) d x d y\\ =\int_{0}^{1} \int_{x^{2}}^{2 x} f(x, y) d y d x$ Thus, the correct answer is false.