91Ó°ÊÓ

The outer integral bounds are given as the interval \(-\pi/4 \leq x \leq \pi/4\), and the inner integral bounds are determined by \(\sin x \leq y \leq \cos x\).

Given the bounds for both outer and inner integrals, we can set up the double integral as follows: $$\int_{-\pi / 4}^{\pi / 4} \int_{\sin x}^{\cos x} 1 \, d y d x$$

To evaluate the inner integral, we integrate the integrand (which is 1) with respect to y: $$\int_{\sin x}^{\cos x} 1 \, d y = [y]_{\sin x}^{\cos x} = (\cos x) - (\sin x)$$

Now, we need to integrate the result of the inner integral with respect to x: $$\int_{-\pi / 4}^{\pi / 4} (\cos x - \sin x) \, d x$$ This integral can be split into two integrals: $$\int_{-\pi / 4}^{\pi / 4} \cos x \, d x - \int_{-\pi / 4}^{\pi / 4} \sin x \, d x$$ Now, we can integrate each term separately: 1) \(\int_{-\pi / 4}^{\pi / 4} \cos x \, d x = [\sin x]_{-\pi / 4}^{\pi / 4} = \sin (\pi / 4) - \sin (-\pi / 4) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}\) 2) \(\int_{-\pi / 4}^{\pi / 4} \sin x \, d x = [-\cos x]_{-\pi / 4}^{\pi / 4} = -\cos (\pi / 4) + \cos (-\pi / 4) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0\) By adding the results of both integrations, we get: $$\sqrt{2} - 0 = \sqrt{2}$$

The value of the given double integral is: $$\int_{-\pi / 4}^{\pi / 4} \int_{\sin x}^{\cos x} d y d x = \sqrt{2}$$