Question

The area enclosed by $y=\sqrt{5-x^2}$ and $y=|x-1|$ is

• A. $\left(\begin{array}{c}\frac{5\pi }{4}-2\end{array}\right)$sq. units
• B. $\frac{5\pi -2}{2}$ sq. units
• C. $\left(\begin{array}{c}\frac{5\pi }{4}-\frac{1}{2}\end{array}\right)$sq. units
• D. $\left(\begin{array}{c}\frac{\pi }{2}-5\end{array}\right)$sq. units

Answer 1

Answer: (C)

$\left(\begin{array}{c}\frac{5\pi }{4}-\frac{1}{2}\end{array}\right)$sq. units

Finding the intersection points

$\sqrt{5-x^2}=|x-1|$

$5-x^2=x^2-2x+1$

$\Rightarrow 2x^2-2x-4=0$

$x=2,-1$

So area $={\int }_{-1}^2(\sqrt{5-x^2}-|x-1|)dx$

$={\int }_{-1}^2\left(\sqrt{5-x^2}\right)dx-{\int }_{-1}^2|x-1|dx$

$={\int }_{-1}^2\left(\sqrt{5-x^2}\right)dx-[{\int }_{-1}^1(1-x)dx+{\int }_1^2(x-1)dx]$

$={[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}]}_{-1}^2-(2-0+\frac{3}{2}-1)$

$=[\sqrt{1}+\frac{5}{2}{\sin }^{-1}\frac{2}{\sqrt{5}}-(-1+\frac{5}{2}{\sin }^{-1}\frac{-1}{\sqrt{5}})]-\frac{5}{2}$

$=2+\frac{5}{2}\times \frac{\pi }{2}-\frac{5}{2}$

$=\frac{5\pi }{4}-\frac{1}{2}$