Area Under Curves Question 6
Question 6 - 29 January - Shift 2
The area of the region
$A=\lbrace (x, y):|\cos x-\sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2} \rbrace$
(1) $1-\frac{3}{\sqrt{2}}+\frac{4}{\sqrt{5}}$
(2) $\sqrt{5}+2 \sqrt{2}-4.5$
(3) $\frac{3}{\sqrt{5}}-\frac{3}{\sqrt{2}}+1$
(4) $\sqrt{5}-2 \sqrt{2}+1$
Show Answer
Answer: (4)
Solution:
Formula: Area between two curves - Area enclosed between two curves intersecting at two different points
$|\cos x-\sin x| \leq y \leq \sin x$
Intersection point of $\cos x-\sin x=\sin x$
$\Rightarrow \tan x=\frac{1}{2}$
Let $\psi=\tan ^{-1} \frac{1}{2}$
So, $\tan \psi=\frac{1}{2}, \sin \psi=\frac{1}{\sqrt{5}}, \cos \psi=\frac{2}{\sqrt{5}}$
$\begin{aligned} \text{ Area } & =\int _{\psi}^{\pi / 2}(\sin x-|\cos x-\sin x|) d x \\ & =\int _{\psi}^{\pi / 4}(\sin x-(\cos x-\sin x)) d x +\int _{\pi / 4}^{\pi / 2}(\sin x-(\sin x-\cos x)) d x \end{aligned}$
Area $=\int _{\psi}^{\pi / 4}(2 \sin x-\cos x) d x+\int _{\pi / 4}^{\pi / 2} \cos x d x$
Area $=[-2 \cos x-\sin x] _{\psi}^{\pi / 4}+[\sin x] _{\pi / 4}^{\pi / 2}$
Area $=-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos \psi+\sin \psi+(1-\frac{1}{\sqrt{2}})$
Area $=-\sqrt{2}-\frac{1}{\sqrt{2}}+2(\frac{2}{\sqrt{5}})+(\frac{1}{\sqrt{5}})+(1-\frac{1}{\sqrt{2}})$
Area $= \sqrt{5}-2 \sqrt{2}+1$
