The area enclosed by the curves $y = \sin x + \cos x$ and $y = |\cos x - \sin x|$ over the interval $[0,\frac{π}{2}]$, is

$2\sqrt{2}(\sqrt{2}-1)$

Let A denote the area enclosed by the given curves. Then,

$A =\int\limits_0^{π/2}|(\sin x + \cos x)-|\cos x - \sin x|| dx$

$∵|\cos x - \sin x|=\left\{\begin{matrix}\cos x - \sin x,&0≤x≤\frac{π}{4}\\-(\cos x - \sin x),&\frac{π}{4}≤x≤\frac{π}{2}\end{matrix}\right.$

$∴\int\limits_0^{π/4}|(\sin x + \cos x)-(\cos x - \sin x)| dx+\int\limits_{π/4}^{π/2}|(\sin x + \cos x)+(\cos x - \sin x)| dx$

$⇒A=\int\limits_0^{π/4}|2\sin x|dx+\int\limits_{π/4}^{π/2}|2\cos x|dx$

$⇒A=2\int\limits_0^{π/4}\sin x\,dx+2\int\limits_{π/4}^{π/2}\cos x\,dx$

$⇒A=\left[-2\cos x\right]_0^{π/4}+\left[2\sin x\right]_{π/4}^{π/2}$

$⇒A=(-2 \cos π/4 + 2) + (2 \sin π/2-2 \sin π/4)$

$⇒A=(2-\sqrt{2})+(2-\sqrt{2})=2(2-\sqrt{2})=2\sqrt{2} (\sqrt{2}-1)$