Question

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

• A. $4(\sqrt{2}-1)$
• B. $2\sqrt{2}(\sqrt{2}-1)$
• C. $2(\sqrt{2}+1)$
• D. $2\sqrt{2}(\sqrt{2}+1)$

Answer 1

The correct option is B $2\sqrt{2}(\sqrt{2}-1)$

The rough graph of $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ suggest the required area is $=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\right(\sin x+\cos x)-|\cos x-\sin x|]dx$
$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}2\sin xdx+\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}2\cos xdx$
$=2\left[\right(-\cos x{)}_0^{\pi /4}+\left(\sin x{)}_{\pi /4}^{\pi /2}\right]=2\sqrt{2}(\sqrt{2}-1)$