Question

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

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

Answer 1

Answer: (B)

Given, $y=\sqrt{5-x^2}\Rightarrow y^2+x^2=5$
and $y=|x-1|$

$\therefore$ Required area
$=\underset{-1}{\overset{2}{\int }}\sqrt{5-x^2}dx-\underset{-1}{\overset{1}{\int }}(1-x)dx-\underset{1}{\overset{2}{\int }}(x-1)dx$
$={\left[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}\right]}_{-1}^2-{\left[x-\frac{x^2}{2}\right]}_{-1}^1-{\left[\frac{x^2}{2}-x\right]}_1^2$
$=\left[1+\frac{5}{2}{\sin }^{-1}\frac{2}{\sqrt{5}}+1+\frac{5}{2}{\sin }^{-1}\frac{1}{\sqrt{5}}\right]$
$-\left[1-\frac{1}{2}-\left(-1-\frac{1}{2}\right)\right]-\left[2-2-\left(\frac{1}{2}-1\right)\right]$
$=2+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}}\right)-\frac{5}{2}$
$=\frac{5}{2}{\sin }^{-1}(1)-\frac{1}{2}=\frac{5\pi }{4}-\frac{1}{2}=\left(\frac{5\pi -2}{4}\right)\text{ sq units }$