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

$\frac{5π-2}{2}$

The graphs of $y=|x-1|$ and $y=\sqrt{5-x^2}$ are shown in Fig and the shaded region is the region bounded
by the two curves.

Let A be the area bounded by the given curves. Then,

$A=\int\limits_{-1}^1(y_2-y_1)dx+\int\limits_{1}^2(y_4-y_3)dx$

$⇒A=\int\limits_{-1}^1(\sqrt{5-x^2}+x-1)dx+\int\limits_{1}^2(\sqrt{5-x^2}-x+1)dx$

$⇒A=\int\limits_{-1}^1\sqrt{5-x^2}dx+\int\limits_{1}^2\sqrt{5-x^2}+\int\limits_{-1}^1(x-1)dx+\int\limits_{1}^2(-x+1)dx$

$⇒A=\int\limits_{-1}^2\sqrt{5-x^2}dx+\left[\frac{x^2}{2}-x\right]_{-1}^1+\left[-\frac{x^2}{2}+x\right]_1^2$

$⇒A=\left[\frac{1}{2}x\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right]_{-1}^2-\frac{5}{2}$

$⇒A=1+\frac{5}{2}\sin^{-1}\frac{2}{\sqrt{5}}+1+\frac{5}{2}\sin^{-1}\frac{1}{\sqrt{5}}-\frac{5}{2}$

$⇒A=-\frac{1}{2}+\frac{5}{2}\sin^{-1}\left(\frac{2}{\sqrt{5}}×\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}}\right)$

$⇒A=-\frac{1}{2}\frac{5}{2}\sin^{-1}(1)=\frac{5π}{4}-\frac{1}{2}=\frac{5π-2}{2}$