The area enclosed by the curves $y=\sin x+\cos x$ and $y=\left|\cos x-\sin x\right|$ and the lines $x=0,x=\frac{\pi }{2}$ is

$A={\int }_0^{\pi /2}\left(\left(\sin x+\cos x\right)-\left|\cos x-\sin x\right|\right)dx$

$A={\int }_0^{\pi /4}\left(\left(\sin x+\cos x\right)-\left(\cos x-\sin x\right)\right)dx+{\int }_{\pi /4}^{\pi /2}\left(\left(\sin x+\cos x\right)-\left(\sin x-\cos x\right)\right)dx$

$A=2{\int }_0^{\pi /4}\sin xdx+2{\int }_{\pi /4}^{\pi /2}\cos xdx$

$A=-2\left(\frac{1}{\sqrt{2}}-1\right)+2\left(1-\frac{1}{\sqrt{2}}\right)$

$A=4-2\sqrt{2}=2\sqrt{2}\left(\sqrt{2}-1\right)$