91Ó°ÊÓ

a. Examining integration with respect to x and y Integrating with respect to x: \(\int_{0}^{1}(y - y^2) dx = \int_{0}^{1}(x - x^2) dx\) Integrating with respect to y: \(\int_{0}^{1}(\sqrt{x} - x) dy = \int_{0}^{1}(\sqrt{y} - y) dy\) We can see that both integrals are possible calculations for the area of the region bounded by the functions \(y=x\) and \(x=y^2\). Therefore, the statement is false. b. Calculating the integral: \(\int_{0}^{\pi / 2}(\cos x-\sin x) d x = [-\sin x - \cos x]_{0}^{\pi / 2} = (-1 - 0) - (0 - 1) = 1\) Since the integral evaluates to 1, the statement is true. c. Comparing the integrals: \(\int_{0}^{1}(x-x^{2}) d x = [\frac{1}{2}x^2 - \frac{1}{3}x^3]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\) \(\int_{0}^{1}(\sqrt{y}-y) dy = [\frac{2}{3}y^{\frac{3}{2}} - \frac{1}{2}y^2]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\) Both integrals evaluate to \(\frac{1}{6}\), so the statement is true.