Using integration, find the area of the region $\left\{\left(x,y\right) \bigg| \left|x-1\right| \leq y \leq \sqrt{5 - x^2} \right\}$
When $x > 1$, $\left|x - 1\right| = x - 1$
When $x < 1$, $\left|x - 1\right| = - \left(x - 1\right) =1 - x$
$\therefore$ $\;$ The given region is
$\left\{\left(x,y\right) \bigg| \left(1 - x\right) \leq y \leq \sqrt{5 - x^2}, \; \left(x - 1\right) \leq y \leq \sqrt{5 - x^2} \right\}$
The equations represented by the inequation $\left(x - 1\right) \leq y \leq \sqrt{5 - x^2}$ $\;$ are
$x - 1 = y$ $\;\;\; \cdots \; (1a)$
$y = \sqrt{5 - x^2}$ $\;\;\; \cdots \; (1b)$
The equation represented by the inequation $\left(1 - x\right) \leq y$ is
$1 - x = y$ $\;\;\; \cdots \; (1c)$
Required area is the shaded region CABC.
$\text{Area} \left(CABC\right) = \text{Area} \left(PABQP\right) - \left[\text{Area} \left(PAC\right) + \text{Area} \left(CBQ\right)\right]$ $\;\;\; \cdots \; (2)$
Let $\text{Area} \left(PABQP\right) = \left|I_1\right|$
$\begin{aligned}
I_1 & = \int \limits_{-1}^{2} y \; dx \hspace{3em} \text{where } \; y = \sqrt{5 - x^2} \\\\
& = \int \limits_{-1}^{2} \sqrt{5 - x^2} \; dx \\\\
& = \int \limits_{-1}^{2} \sqrt{\left(\sqrt{5}\right)^2 - \left(x\right)^2} \; dx \\\\
& \left[\text{Note: } \int \limits_{p}^{q} \sqrt{a^2 - x^2} \; dx = \left[\dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right)\right]_{p}^{q}\right] \\\\
& = \left[\dfrac{x}{2} \times \sqrt{5 - x^2} + \dfrac{5}{2} \sin^{-1} \left(\dfrac{x}{\sqrt{5}}\right)\right]_{-1}^{2} \\\\
& = \dfrac{2}{2} \times \sqrt{5 - 4} + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}}\right) - \left[\dfrac{-1}{2} \times \sqrt{5 - 1} + \dfrac{5}{2} \sin^{-1} \left(\dfrac{-1}{\sqrt{5}}\right)\right] \\\\
& = 1 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}}\right) + \dfrac{1}{2} \times 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right) \\\\
& = 2 + \dfrac{5}{2} \left[\sin^{-1} \left(\dfrac{2}{\sqrt{5}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right)\right] \\\\
& \left[\text{Note: } \sin^{-1}\left(x\right) + \sin^{-1} \left(y\right) = \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right)\right] \\\\
& = 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}} \times \sqrt{1 - \dfrac{1}{5}} + \dfrac{1}{\sqrt{5}} \times \sqrt{1 - \dfrac{4}{5}}\right) \\\\
& = 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}} \times \dfrac{2}{\sqrt{5}} + \dfrac{1}{\sqrt{5}} \times \dfrac{1}{\sqrt{5}}\right) \\\\
& = 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{4}{5} + \dfrac{1}{5}\right) \\\\
& = 2 + \dfrac{5}{2} \sin^{-1} \left(1\right) \\\\
& = 2 + \dfrac{5}{2} \times \dfrac{\pi}{2} = 2 + \dfrac{5 \pi}{4}
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(PABQP\right) = 2 + \dfrac{5 \pi}{4}$ sq units $\;\;\; \cdots \; (3)$
Let $\text{Area} \left(PAC\right) = \left|I_2\right|$
$\begin{aligned}
I_2 & = \int \limits_{-1}^{1} y \; dx \hspace{2em} \text{ where } \; y = 1-x \\\\
& = \int \limits_{-1}^{1} \left(1 - x\right) \; dx \\\\
& = \left[x - \dfrac{x^2}{2}\right]_{-1}^{1} \\\\
& = 1 - \dfrac{1}{2} - \left[-1 - \dfrac{1}{2}\right] = 2
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(PAC\right) = 2$ sq units $\;\;\; \cdots \; (4)$
Let $\text{Area} \left(CBQ\right) = \left|I_3\right|$
$\begin{aligned}
I_3 & = \int \limits_{1}^{2} y \; dx \hspace{2em} \text{ where } \; y = x-1 \\\\
& = \int \limits_{1}^{2} \left(x - 1\right) \; dx \\\\
& = \left[\dfrac{x^2}{2} - x\right]_{1}^{2} \\\\
& = \dfrac{4}{2} - 2 - \dfrac{1}{2} + 1 = \dfrac{1}{2}
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(CBQ\right) = \dfrac{1}{2}$ sq units $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ In view of equations $(3)$, $(4)$ and $(5)$, equation $(2)$ becomes
$\text{Area} \left(CABC\right) = 2 + \dfrac{5 \pi}{4} - 2 - \dfrac{1}{2} = \dfrac{5 \pi}{4} - \dfrac{1}{2}$ sq units