CBSE Class 12 · 2010 · Area Under CurvesLong Answer

The region is bounded above by the circle $y=\sqrt{5-x^2}$ (radius $\sqrt{5}$) and below by the V-shaped graph $y=|x-1|$.
Intersection: $x-1=\sqrt{5-x^2}\Rightarrow 2x^2-2x-4=0\Rightarrow (x-2)(x+1)=0$, so $x=-1$ and $x=2$.

The first integral $=[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}{]}_{-1}^2=2+\frac{5}{2}({\sin }^{-1}\frac{2}{\sqrt{5}}+{\sin }^{-1}\frac{1}{\sqrt{5}})=2+\frac{5}{2}\cdot \frac{\pi }{2},$ since ${\sin }^{-1}\frac{2}{\sqrt{5}}+{\sin }^{-1}\frac{1}{\sqrt{5}}=\frac{\pi }{2}$.
The two linear pieces give $2+\frac{1}{2}=\frac{5}{2}$ of area under $|x-1|$.