The area of the region A=(x,y): | x- x| y x, 0 x 2 is:

Question

The area of the region $A=\{(x,y): |\cos x-\sin x|\le y\le \sin x,\ 0\le x\le \dfrac{\pi}{2}\}$ is:

Accepted Answer

$$
\cos x-\sin x| \leq y \leq \sin x
$$

Intersection point of $\cos x-\sin x=\sin x$

$$
\Rightarrow \tan x=\frac{1}{2}
$$

Let $\psi=\tan ^{-1} \frac{1}{2}$

So, $\tan \psi=\frac{1}{2}, \sin \psi=\frac{1}{\sqrt{5}}, \cos \psi=\frac{2}{\sqrt{5}}$

Area $$ = \int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{{\pi  \over 4}} {(\sin x - (\cos x - \sin x))dx + \int\limits_{{\pi  \over 4}}^{{\pi  \over 2}} {(\sin x - (\sin x - \cos x))dx} } $$

$$ =\int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}(2 \sin x-\cos x) d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x d x $$

$$ =[-2 \cos x-\sin x]_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2} $$

$$ =-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos ({{{\tan }^{ - 1}}{1 \over 2}}) +\sin ({{{\tan }^{ - 1}}{1 \over 2}})+\left(1-\frac{1}{\sqrt{2}}\right)$$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)+1-\frac{1}{\sqrt{2}}$

$$ = \sqrt 5  - 2\sqrt 2  + 1$$

Solutions · 1