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

• A

  $4(\sqrt{2}-1)$
• Correct answer
  B

  $2 \sqrt{2}(\sqrt{2}-1)$
• C

  $2(\sqrt{2}+1)$
• D

  $2 \sqrt{2}(\sqrt{2}+1)$

Solution

Given, $y=\sin x+\cos x, x \in[0, \pi / 2]$ $\frac{d y}{d x}=\cos x-\sin x$ $\begin{aligned} y & =|\cos x-\sin x| \\ & =\left[\begin{array}{cc}\cos x-\sin x & x \in[0, \pi / 4] \\ \sin x-\cos x & x \in[\pi / 4, \pi / 2]\end{array}\right]\end{aligned}$
$\therefore$ Required area $\begin{aligned} &=\int_0^{\pi / 4}|(\sin x+\cos x)-(\cos x-\sin x)| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x \end{aligned}$ $=\int_0^{\pi / 4}|2 \sin x| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x$ $=2(-\cos x)_0^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2}$ $=2\left[-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\right]$
$\begin{aligned}= & 2\left(2-\frac{2}{\sqrt{2}}\right) \\ = & 2(2-\sqrt{2}) \\ = & 4-2 \sqrt{2} \\ = & 2 \sqrt{2}(\sqrt{2}-1)\end{aligned}$

Hence, the correct answer is option: 'B'.