Mathematics - Area Under Curves Question with Solution | TestHub

Given, $y=\sin x+\cos x,x\in [0,\pi /2]$ $\frac{dy}{dx}=\cos x-\sin x$ $\begin{array}{rl}y& =|\cos x-\sin x|\\ & =\left[\begin{array}{cc}\cos x-\sin x& x\in [0,\pi /4]\\ \sin x-\cos x& x\in [\pi /4,\pi /2]\end{array}\right]\end{array}$
$\therefore$ Required area $\begin{array}{rl}& ={\int }_0^{\pi /4}|(\sin x+\cos x)-(\cos x-\sin x)|dx+{\int }_{\pi /4}^{\pi /2}|2\cos x|dx\end{array}$ $={\int }_0^{\pi /4}|2\sin x|dx+{\int }_{\pi /4}^{\pi /2}|2\cos x|dx$ $=2(-\cos x{)}_0^{\pi /4}+2(\sin x{)}_{\pi /4}^{\pi /2}$ $=2\left[-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\right]$
$\begin{array}{rl}=& 2\left(2-\frac{2}{\sqrt{2}}\right)\\ =& 2(2-\sqrt{2})\\ =& 4-2\sqrt{2}\\ =& 2\sqrt{2}(\sqrt{2}-1)\end{array}$