Mathematics - Area Under Curves Question with Solution | TestHub

The graphs of $y=|x-1|$ and $y=\sqrt{5-x^2}$ are shown in the figure and the shaded region is the required region bounded by the two curves.Let $A$ be the area bounded by the given curves. Then $$\begin{array}{rl}A& ={\int }_{-1}^1\left(\sqrt{5-x^2}+x-1\right)dx+{\int }_1^2\left(\sqrt{5-x^2}-x+1\right)dx\\ & ={\int }_{-1}^1\sqrt{5-x^2}dx+{\int }_1^2\sqrt{5-x^2}dx+{\int }_{-1}^1(x-1)dx+{\int }_1^2(-x+1)dx\\ & ={\int }_{-1}^2\sqrt{5-x^2}dx+{\left[\frac{x^2}{2}-x\right]}_{-1}^1+{\left[-\frac{x^2}{2}+x\right]}_1^2\\ & ={\left[\frac{1}{2}x\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}\right]}_{-1}^2-\frac{5}{2}\\ & =1+\frac{5}{2}{\sin }^{-1}\frac{2}{\sqrt{5}}+1+\frac{5}{2}{\sin }^{-1}\frac{1}{\sqrt{5}}-\frac{5}{2}\\ & =-\frac{1}{2}+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\times \sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}}\right)\\ & =-\frac{1}{2}+\frac{5}{2}{\sin }^{-1}(1)=\frac{5\pi }{4}-\frac{1}{2}\mathrm{squnits}\end{array}$$