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  • The area of the region  <img alt="A=\left\{(x, y):|\cos x-\sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\right\}" src="https://entrancecorner.oncodecogs.com/gif.latex?A%3D%5Cleft%5C%7

The area of the region  A=\left\{(x, y):|\cos x-\sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\right\} is

Option: 1

\sqrt{5}+2 \sqrt{2}-4.5


Option: 2

1-\frac{3}{\sqrt{2}}+\frac{4}{\sqrt{5}}


Option: 3

\frac{3}{\sqrt{5}}-\frac{3}{\sqrt{2}}+1


Option: 4

\sqrt{5}-2 \sqrt{2}+1


Answers (1)

best_answer

                                          \mathrm{A}= area under the curve
                                                                                                 y=\sin x \&  above the curve |\cos x-\sin x|
                                                                                              $$ A=\int_0^{\pi / 2}(\sin x-|\cos x-\sin x|) d x

\text { When } 0 \text { to } \frac{\pi}{4} \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \text { when } \frac{\pi}{4} \text { to } \frac{\pi}{2}

|\cos x-\sin x|=\cos x-\sin x \, \, \, \, \, \, \, \, \, \, \quad|\cos x-\sin x|=\sin x-\cos x                                                                                    
                                                                                                                   
 \sin x=\cos x-\sin x\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad \sin x=\sin x-\cos x

2 \sin x=\cos x \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad \cos x=0
\tan x=\frac{1}{2} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad x=\frac{\pi}{2}

x=\tan ^{-1}\left(\frac{1}{2}\right)

A=\int_{\tan ^{-1}(1 / 2)}^{\pi / 4}\{\sin x-(\cos x-\sin x)\} d x+\int_{\pi / 4}^{\pi / 2}\{\sin x+(\cos x-\sin x)\} d x

\begin{aligned} & \Rightarrow \int_{\tan ^{-1}\left(\frac{1}{2}\right)}^{\pi / 4}(2 \sin x-\cos x) d x+\int_{\pi / 4}^{\pi / 2} \cos x d x \\ & \Rightarrow(-2 \cos x-\sin x)_{\tan ^{-1}\left(\frac{1}{2}\right)}^{\pi / 4}+(\sin x)_{\pi / 4}^{\pi / 2} \\ & \Rightarrow\left(-2 \times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-\left\{-2 \cos \left(\tan ^{-1} \frac{1}{2}\right)-\sin \left(\tan ^{-1} \frac{1}{2}\right)\right\}+\left(1-\frac{1}{\sqrt{2}}\right) \\ & \Rightarrow-\frac{3}{\sqrt{2}}+2 \times \frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}+1-\frac{1}{\sqrt{2}}=\frac{-4}{\sqrt{2}}+\frac{5}{\sqrt{5}}+1=-2 \sqrt{2}+\sqrt{5}+1 \end{aligned}

Posted by

Ritika Kankaria

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