Question

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

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

Answer 1

The correct option is B

$2\sqrt{2}\left(\sqrt{2}-1\right)$

Explanation for the correct answer:

Finding the area enclosed by the curves:

Consider the given curve as,

$y_1=\sin x+\cos x$

$y_2=\left|\cos x-\sin x\right|$

The given interval is $\left[0,\frac{\mathrm{\pi }}{2}\right]$

Let,

$$\begin{gathered} I_1={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sin x+\cos x\right)dx \\ {I}_1={\left[-\cos x+\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ {I}_1=\left(-c\mathrm{os}\frac{\mathrm{\pi }}{2}+\sin \frac{\mathrm{\pi }}{2}-\left(-\cos 0+\sin 0\right)\right) \\ {I}_1=\left(0+1+1-0\right) \\ {I}_1=2 \end{gathered}$$

Now

$y_2=\left|\cos x+\sin x\right|=\left\{\begin{array}{ll}\cos x-\sin x,& 0\le x\le \frac{\mathrm{\pi }}{4}\\ -\left(\cos x-\sin x\right)& \frac{\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{2}\end{array}\right.$

Let,

$$\begin{gathered} I_2={\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\cos x-\sin x\right|dx \\ {I}_2={\int }_0^{\frac{\mathrm{\pi }}{4}}\left(\cos x-\sin x\right)dx+{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}-\left(\cos x-\sin x\right)dx \\ ={\left[\sin x+\cos x\right]}_0^{\frac{\mathrm{\pi }}{4}}+{\left[-\sin x+\cos x\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ =\sin \frac{\mathrm{\pi }}{4}+\cos \frac{\mathrm{\pi }}{4}-\sin 0-\cos 0+\left(-\sin \frac{\mathrm{\pi }}{2}-\cos \frac{\mathrm{\pi }}{2}+\sin \frac{\mathrm{\pi }}{4}+\cos \frac{\mathrm{\pi }}{4}\right) \\ =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1-1-0+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ =\frac{4}{\sqrt{2}}-2 \\ =2\sqrt{2}-2 \end{gathered}$$

Then, the required area is

$$\begin{gathered} I_1-I_2=2-\left(2\sqrt{2}-2\right) \\ =4-2\sqrt{2} \\ =2\sqrt{2}\left(\sqrt{2}-1\right) \end{gathered}$$

Hence, the area enclosed by the given curve is equal to $2\sqrt{2}\left(\sqrt{2}-1\right)$

Hence, option (B) is the correct answer.