Question:
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:
Correct Option: 1
Solution:
$A=\int_{0}^{\pi / 2}((\sin x+\cos x)-|\cos x-\sin x|) d x$
$A=\int_{0}^{\pi / 2}((\sin x+\cos x)-(\cos x-\sin x)) d x$
$+\int_{\pi / 4}^{\pi / 2}((\sin x+\cos x)-(\sin x-\cos x)) d x$
$A=2 \int_{0}^{\pi / 2} \sin x d x+2 \int_{\pi / 4}^{\pi / 2} \cos x d x$
$A=-2\left(\frac{1}{\sqrt{2}}-1\right)+2\left(1-\frac{1}{\sqrt{2}}\right)$
$A=4-2 \sqrt{2}=2 \sqrt{2}(\sqrt{2}-1)$
Option (1)