The area, enclosed by the curves


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

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

  2. $2(\sqrt{2}+1)$

  3. $4(\sqrt{2}-1)$

  4. $2 \sqrt{2}(\sqrt{2}+1)$

Correct Option: 1


$A=\int_{0}^{\pi / 2}((\sin x+\cos x)-|\cos x-\sin x|) d x$

$A=\int_{0}^{\pi / 2}((\sin x+\cos x)-(\cos x-\sin x)) d x$

$+\int_{\pi / 4}^{\pi / 2}((\sin x+\cos x)-(\sin x-\cos x)) d x$

$A=2 \int_{0}^{\pi / 2} \sin x d x+2 \int_{\pi / 4}^{\pi / 2} \cos x d x$


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

Option (1)

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