Explanation

(b)
$Given,\ \ y = \sqrt{5 - x^{2}}\ and\ \ y = |x - 1|$
$or\ \ y^{2} + x^{2} = 5$
$and\ \ y = |x - 1|$
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$\therefore$ Required area
$= \int_{- 1}^{2}\sqrt{5 - x^{2}}dx - \int_{- 1}^{1}{(1 - x)}dx - \int_{1}^{2}{(x - 1)}dx$
$= \left\lbrack \frac{x}{2}\sqrt{5 - x^{2}} + \frac{5}{2}\sin^{- 1}\frac{x}{\sqrt{5}} \right\rbrack_{- 1}^{2} - \left\lbrack x - \frac{x^{2}}{2} \right\rbrack_{- 1}^{1} - \left\lbrack \frac{x^{2}}{2} - x \right\rbrack_{1}^{2}$
$= \left\lbrack 1 + \frac{5}{2}\sin^{- 1}\frac{2}{\sqrt{5}} + 1 + \frac{5}{2}\sin^{- 1}\frac{1}{\sqrt{5}} \right\rbrack - \left\lbrack 1 - \frac{1}{2} - \left( - 1 - \frac{1}{2} \right) \right\rbrack - \left\lbrack 2 - 2 - \left( \frac{1}{2} - 1 \right) \right\rbrack$
$= 2 + \frac{5}{2}\sin^{- 1}\left( \frac{2}{\sqrt{5}}\sqrt{1 - \frac{1}{5}} + \frac{1}{\sqrt{5}}\sqrt{1 - \frac{4}{5}} \right) - \frac{5}{2}$
$= \frac{5}{2}\sin^{- 1}{(1)} = \frac{5\pi}{4} - \frac{1}{2} = \left( \frac{5\pi - 2}{4} \right)sq\ unit$