Question

The area enclosed by the curves $y=sinx+$cosx and $y=|cosx-$sinx $|$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 $2\sqrt{2}(\sqrt{2}-1)$

$\cos x>\sin x,\forall xϵ(0,\frac{\pi }{4})$, and $\cos x<\sin x,\forall xϵ(\frac{\pi }{4},\frac{\pi }{2})$
$y_1=\sin x+\cos x$
$y_2=|\cos x-\sin x|$
$\Rightarrow$Area

$={\int }_0^{\pi /2}(y_1-y_2)dx$

$=$ ${\int }_0^{\pi /4}$ $\left(\right(sinx+cosx)-(cosx-sinx\left)\right)dx$ $+{\int }_{\pi /4}^{\pi /2}\left(\right(sinx+cosx)-(sinx-cosx\left)\right)dx$

$=4-2\sqrt{2}$