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
$=\int\limits_{-1}^2 \sqrt{5-x^2} d x-\int\limits_{-1}^1(1-x) d x-\int\limits_1^2(x-1) d x $
$ =\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}$
$=\frac{5}{2} \sin ^{-1}(1)-\frac{1}{2}=\frac{5 \pi}{4}-\frac{1}{2}=\left(\frac{5 \pi-2}{4}\right) \text { sq units }$