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:

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

Answer 1

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

$y=|\cos x-\sin x|=|\sin x-\cos x|$
$=\{\begin{array}{cc}\cos x-\sin x& 0\le x\le \frac{\pi }{4}\\ \sin x-\cos x& \frac{\pi }{4}\le x\le \frac{\pi }{2}\end{array}$
$A=\underset{0}{\overset{\pi /4}{\int }}(\sin x+\cos x)-(\cos x-\sin x)dx+\underset{\pi /4}{\overset{\pi /2}{\int }}(\sin x+\cos x)-(\sin x-\cos x)dx$
$=2\underset{0}{\overset{\pi /4}{\int }}\sin xdx+2\underset{\pi /2}{\overset{\pi /4}{\int }}\cos xdx=2(-\cos x){|}_0^{\pi /4}+2\sin x{|}_{\pi /4}^{\pi /2}$
$=2(1-\frac{1}{\sqrt{2}})+2(1-\frac{1}{\sqrt{2}})$
$=4(1-\frac{1}{\sqrt{2}})$
$=2\sqrt{2}(\sqrt{2}-1)$