Question

Using integration find the area of the region: $\left\{\left(x,y\right):\left|x-1\right|\le y\le \sqrt{5-x^2}\right\}$.

Answer 1

$$\begin{gathered} \left|x-1\right|\le y\le \sqrt{5-x^2} \\ \left|x-1\right|=\sqrt{5-x^2} \\ x=2,-1 \\ \mathrm{A}={\int }_{-1}^2\left(\sqrt{5-x^2}-\left|x-1\right|\right)dx \\ ={\int }_{-1}^2\sqrt{5-x^2}+{\int }_{-1}^1\left(x-1\right)dx+{\int }_1^2\left(1-x\right)dx \\ ={\left[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)\right]}_{-1}^2+{\left[\frac{x^2}{2}-x\right]}_{-1}^1+{\left[x-\frac{x^2}{2}\right]}_1^2 \\ =\frac{5}{2}\left({\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)+{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+\frac{1}{2} \end{gathered}$$