Question

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

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

Answer 1

The correct option is B $\frac{(5\pi -2)}{4}$ square units

$y=|x-1|\Rightarrow y=x-1$ when $x>1$ and $y=-(x-1)$ when $x<1$
and $y=\sqrt{5-x^2}$ is the semicircle above x-axis
Finding intersection points of $y=\sqrt{5-x^2}$ and $y=x-1$ which gives $(2,1)$ and $y=\sqrt{5-x^2}$ with $y=-x+1$ which gives $(-1,2)$
$\therefore$ Required bounded area
$={\int }_{-1}^2\sqrt{5-x^2}dx-{\int }_{-1}^2\left|\right(x-1\left)\right|dx={[\frac{x}{2}+\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}]}_{-1}^2-[{\int }_{-1}^1(-x+1)dx+{\int }_1^2(x-1)dx]$
$=\frac{5\pi }{4}-\frac{1}{2}sq.units$