Question

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

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

Answer 1

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

$\underset{0}{\overset{\pi /2}{\int }}(\sin x+\cos x)dx-[\underset{0}{\overset{\pi /4}{\int }}(\cos x-\sin x)dx+\underset{\pi /4}{\overset{\pi /2}{\int }}(\sin x-\cos x)dx]$
$=-{\left|\cos x\right|}_0^{\pi /2}+{\left|\sin x\right|}_0^{\pi /2}-[{\left|\sin x\right|}_0^{\pi /4}+{\left|\cos x\right|}_0^{\pi /4}-{\left|\cos x\right|}_{\pi /4}^{\pi /2}-{\left|\sin x\right|}_{\pi /4}^{\pi /2}]$
$=-(0-1)+(1-0)-[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1-(0-\frac{1}{\sqrt{2}})-(1-\frac{1}{\sqrt{2}})]$

$=2-[\sqrt{2}-1+\frac{1}{\sqrt{2}}-1+\frac{1}{\sqrt{2}}]$
$=2-[2\sqrt{2}-2]$
$=4-2\sqrt{2}=2\sqrt{2}(\sqrt{2}-1)$