Question

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

Answer 1

Given:

Curves are $y=\sqrt{5-x^2}$ and $y=|x-1|$

$y=\sqrt{5-x^2}$ represent a semicircle with radius $\sqrt{5}$

and center $(0,0)$ and y is positive so semicircle will be above x axis.

$y=|x-1| can be written as y=x-1 when x>1 and 1-x when x<1. (represented by graph.

Solving $y=\sqrt{5-x^2}$ and $y=|x-1|$

We get two point of intersection $A(-1,2)$ and $B(2,1).$

From the figure area of shaded part will be

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

$A={\left[\begin{array}{c}\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}(\frac{x}{\sqrt{5}})\end{array}\right]}_{-1}^2-{\left[\begin{array}{c}x-\frac{x^2}{2}\end{array}\right]}_{-1}^1-{\left[\begin{array}{c}\frac{x^2}{2}-x\end{array}\right]}_1^2$

On applying limits,

$1+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)+1-\frac{5}{2}{\sin }^{-1}\left(\frac{-1}{\sqrt{5}}\right)-1+\frac{1}{2}-1-\frac{1}{2}-2+2+\frac{1}{2}-1$

$\Rightarrow A=\frac{5}{2}({\sin }^{-1}\frac{2}{\sqrt{5}}+{\sin }^{-1}\frac{1}{\sqrt{5}})-\frac{1}{2}$

We know that ${\sin }^{-1}\frac{1}{\sqrt{5}}={\cos }^{-1}\frac{2}{\sqrt{5}}$

$\Rightarrow A=\frac{5}{2}(si{n}^{-1}\frac{2}{\sqrt{5}}+co{s}^{-1}\frac{2}{\sqrt{5}})-\frac{1}{2}$

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

$\Rightarrow \frac{5}{2}.\frac{\pi }{2}-\frac{1}{2}=\frac{5\pi -2}{4}sq.units.$