Question

Sketch the region bounded by the curves $y=\sqrt{5-x^2}$ and $y=|x-1|$ and find its area using intergration.

Answer 1

Given curves are
$y=\sqrt{5-x^2}⟶1$
$y=|x-1|$
$y=\sqrt{5-x^2}$ represents a semi circle of radius $\sqrt{5}$ and having center at $(0,0)$
(Image)
To find point of intersection
For point A
$y^2=5-x^2$ (From $1$)
$y=x-1$
$x^2+{(x-1)}^2=5\to x=2,y=1$
Point $A(2,1)$
For point B,
$y^2=5-x^2$ (From $1$)
$y=1-x$
$x^2+{(1-x)}^2=5\to x=-1,y=2$
Point $B(-1,2)$
$A={\int }_{-1}^2\sqrt{5-x^2}dx-{\int }_{-1}^1(1-x)dx-{\int }_1^2(x-1)dx$
On integrating:
$A={[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)]}_{-1}^2-{[x-\frac{x^2}{2}]}_{-1}^1-{[\frac{x^2}{2}-x]}_1^2$
$\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)+1-\frac{5}{2}{\sin }^{-1}\left(\frac{-1}{\sqrt{5}}\right)-1+\frac{1}{2}-1-\frac{1}{2}-2+2+\frac{1}{2}-1$
$A=\frac{5}{2}({\sin }^{-1}\frac{2}{\sqrt{5}}+{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right))-\frac{1}{2}$
${\sin }^{-1}\frac{1}{\sqrt{5}}={\cos }^{-1}\frac{2}{\sqrt{5}}+{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$
$\therefore \frac{5}{2}.\frac{\pi }{2}-\frac{1}{2}=\left(\frac{5x-2}{2}\right)$sq. units
$=3.43$ sq. units