Question:

The area enclosed by the curves $y = \sin x + \cos x$ and $y = |\cos x - \sin x|$ over the interval $[0, \pi/2]$ is determined by integrating the difference between the upper and lower curves. First, analyze the absolute value function $y = |\cos x - \sin x|$ over the given interval. For $x \in [0, \pi/4]$, $\cos x \ge \sin x$, so $|\cos x - \sin x| = \cos x - \sin x$. For $x \in [\pi/4, \pi/2]$, $\sin x \ge \cos x$, so $|\cos x - \sin x| = \sin x - \cos x$. Next, determine which function is the upper curve. Let $f(x) = \sin x + \cos x$ and $g(x) = |\cos x - \sin x|$. We compare $f(x)^2$ and $g(x)^2$: $f(x)^2 = (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin(2x)$. $g(x)^2 = (\cos x - \sin x)^2 = \cos^2 x + \sin^2 x - 2\sin x \cos x = 1 - \sin(2x)$. For $x \in [0, \pi/2]$, $\sin(2x) \ge 0$. Thus, $f(x)^2 \ge 1$ and $g(x)^2 \le 1$. Since both $f(x)$ and $g(x)$ are non-negative on $[0, \pi/2]$, it follows that $f(x) \ge g(x)$ for all $x \in [0, \pi/2]$. Therefore, the area is given by $\int_0^{\pi/2} ((\sin x + \cos x) - |\cos x - \sin x|) dx$. The integral for the area $A$ is split into two parts based on the definition of the absolute value: $$A = \int_0^{\pi/4} [(\sin x + \cos x) - (\cos x - \sin x)] dx + \int_{\pi/4}^{\pi/2} [(\sin x + \cos x) - (\sin x - \cos x)] dx$$ Simplify the integrands: For the first integral: $(\sin x + \cos x) - (\cos x - \sin x) = \sin x + \cos x - \cos x + \sin x = 2\sin x$. For the second integral: $(\sin x + \cos x) - (\sin x - \cos x) = \sin x + \cos x - \sin x + \cos x = 2\cos x$. Substitute the simplified integrands back into the area formula: $$A = \int_0^{\pi/4} 2\sin x \,dx + \int_{\pi/4}^{\pi/2} 2\cos x \,dx$$ Evaluate the integrals: $$A = [-2\cos x]_0^{\pi/4} + [2\sin x]_{\pi/4}^{\pi/2}$$ Substitute the limits of integration: $$A = (-2\cos(\pi/4) - (-2\cos(0))) + (2\sin(\pi/2) - 2\sin(\pi/4))$$ $$A = \left(-2 \cdot \frac{\sqrt{2}}{2} - (-2 \cdot 1)\right) + \left(2 \cdot 1 - 2 \cdot \frac{\sqrt{2}}{2}\right)$$ $$A = (-\sqrt{2} + 2) + (2 - \sqrt{2})$$ $$A = 2 - \sqrt{2} + 2 - \sqrt{2}$$ $$A = 4 - 2\sqrt{2}$$