Question

Using integration, find the area of the region bounded by the curves $y=\sqrt{5-x^2}$ and $y=|x-1|$

Answer 1

We have $y=\sqrt{5-x^2}\dots \left(i\right)$ and $y=|x-1|=\{\begin{array}{ccc}x-1,& if& x\ge 1\\ 1-x,& if& x<1\end{array}\dots \left(ii\right)$

Solving (i) and (ii), $|x-1{|}^2=[\sqrt{5-x^2}{]}^2$
$\begin{array}{ccc}& & \\ \Rightarrow x^2-2x+1=5-x^2& \Rightarrow x^2-x-2=0& y=|x-1|\\ \Rightarrow (x+1)(x-2)=0& \therefore x=-1,2& \left[1\right]\end{array}$

$\therefore$ Required area $={\int }_{-1}^2{y}_{i}dx-{\int }_{-1}^2{y}_{ii}dx\left[2\right]$

$\Rightarrow ={\int }_{-1}^2\sqrt{5-x^2}dx-{\int }_{-1}^2|x-1|dx\left[2\right]$

$\Rightarrow ={\int }_{-1}^2\sqrt{5-x^2}dx-[{\int }_{-1}^1|x-1|dx+{\int }_1^2|x-1|dx]$

$\Rightarrow ={\int }_{-1}^2=\sqrt{5-x^2}dx-{\int }_{-1}^1-(x-1)dx-{\int }_1^2(x-1)dx$

$\Rightarrow ={[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}si{n}^{-1}\frac{x}{\sqrt{5}}]}_{-1}^2+{\left[\frac{(x-1{)}^2}{2}\right]}_{-1}^1-{\left[\frac{(x-1{)}^2}{2}\right]}_1^2$

$\Rightarrow ={[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}si{n}^{-1}\frac{x}{\sqrt{5}}]}_{-1}^2{\left[\frac{(x-1{)}^2}{2}\right]}_{-1}^1-{\left[\frac{(x-1{)}^2}{2}\right]}_1^2$

$\Rightarrow [1+\frac{5}{2}si{n}^{-1}\frac{2}{\sqrt{5}}]-[-1-\frac{5}{2}si{n}^{-1}\frac{1}{\sqrt{5}}]+0-2-\frac{1}{2}+0$

$\Rightarrow \frac{5}{2}[si{n}^{-1}\frac{2}{\sqrt{5}}+si{n}^{-1}\frac{1}{\sqrt{5}}]-\frac{1}{2}$ sq. units

$\Rightarrow \frac{5}{2}\left[si{n}^{-1}(\frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\frac{4}{5})\right]-\frac{1}{2}$

$\Rightarrow \frac{5}{2}\left[si{n}^{-1}(\frac{4}{5}+\frac{1}{5})\right]-\frac{1}{2}=\frac{5}{2}\times \frac{\pi }{2}-\frac{1}{2}=(\frac{5\pi }{4}-\frac{1}{2})$ sq. units.