The area enclosed by the curves $y = \sin x + \cos x$ and $y = |\cos x − \sin x|$ over the interval [0, π/2] is $(1) 4(√2 − 1) (2) 2√2(√2 − 1) (3) 2(√2 + 1) (4) 2√2(√2 + 1)$

$y_1 = \sin x + \cos x, \qquad y_2 = |\cos x - \sin x|$ $x \in \left[0, \tfrac{\pi}{2}\right]$ $\text{Remove the modulus, find upper/lower, then } A = \int_0^{\pi/2} (\text{upper} - \text{lower})\,dx$ $|\cos x - \sin x| = \begin{cases} \cos x - \sin x & x \in [0,\tfrac{\pi}{4}] \\[2pt] \sin x - \cos x & x \in [\tfrac{\pi}{4},\tfrac{\pi}{2}] \end{cases}$ $y_1 - y_2 = 2\sin x \geq 0 \quad \left(0 \leq x \leq \tfrac{\pi}{4}\right)$ $y_1 - y_2 = 2\cos x \geq 0 \quad \left(\tfrac{\pi}{4} \leq x \leq \tfrac{\pi}{2}\right)$ $\Rightarrow y_1 \text{ is on top throughout}$ $A = \int_0^{\pi/4} 2\sin x\,dx + \int_{\pi/4}^{\pi/2} 2\cos x\,dx$ $\int_0^{\pi/4} 2\sin x\,dx = [-2\cos x]_0^{\pi/4} = -\tfrac{2}{\sqrt{2}} + 2 = 2\left(1 - \tfrac{1}{\sqrt{2}}\right)$ $\int_{\pi/4}^{\pi/2} 2\cos x\,dx = [2\sin x]_{\pi/4}^{\pi/2} = 2 - \tfrac{2}{\sqrt{2}} = 2\left(1 - \tfrac{1}{\sqrt{2}}\right)$ $A = 4\left(1 - \tfrac{1}{\sqrt{2}}\right) = 4 - 2\sqrt{2} = 2\sqrt{2}(\sqrt{2} - 1)$ $\Rightarrow \text{option (2)}$ $A = 2\sqrt{2}(\sqrt{2} - 1) \text{ square units}$