Question

The area bounded by the curves $y=\sin x+\cos x$ 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 $2\sqrt{2}(\sqrt{2}-1)$

Clearly, $\sin x+\cos x\ge |\cos x-\sin x|$ in $[0,\frac{\pi }{2}]$

Required area $=\underset{0}{\overset{\pi /2}{\int }}(\sin x+\cos x)dx-\underset{0}{\overset{\pi /2}{\int }}|\cos x-\sin x|dx$

$=\underset{0}{\overset{\pi /2}{\int }}(\sin x+\cos x)dx-\left[\underset{0}{\overset{\pi /4}{\int }}\right(\cos x-\sin x)dx+\underset{\pi /4}{\overset{\pi /2}{\int }}(\sin x-\cos x\left)dx\right]$

$=(1+1)-[(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)+(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)]$

$=2-\left[\right(\sqrt{2}-1)+(\sqrt{2}-1\left)\right]$

$=4-2\sqrt{2}$

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