Question

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

Answer 1

Consider the given equation

${\displaystyle y=\sqrt{5-x^2}}$

This equation represents a semicircle with centre at

the origin and radius = sqrt5 units

Given that the region is bounded by the above
semicircle and the line y = |x-1|
Let us find the point of intersection of the
given curve meets the line y= |x - 1|

${\displaystyle \Rightarrow \sqrt{5-x^2}=|x-1|}$

Squaring both the sides, we have,

${\displaystyle 5-x^2={|x-1|}^2}$

${\displaystyle \Rightarrow 5-x^2=x^2+1-2x}$

${\displaystyle \Rightarrow 2x^2-2x-5+1=0}$

${\displaystyle \Rightarrow 2x^2-2x-4=0}$

${\displaystyle \Rightarrow x^2-x-2=0}$

${\displaystyle \Rightarrow x^2-2x+x-2=0}$

${\displaystyle \Rightarrow x(x-2)+1(x-2)=0}$

${\displaystyle \Rightarrow (x+1)(x-2)=0}$

${\displaystyle \Rightarrow x=-1,x=2}$

When x = -1,y = 2
When x = 2,y = 1
Consider the following figure.

Thus the intersection points are ( -1,2) and (2,1)
Consider the following sketch of the bounded region.

Required Area, A= ${\displaystyle {\int }_{-1}^2(y_2-y_1)dx}$

=${\displaystyle ={\int }_{-1}^1[\sqrt{5-x^2}+(x-1)]dx+{\int }_1^2[\sqrt{5-x^2}-(x-1)]dx}$

${\displaystyle ={\int }_{-1}^1\sqrt{5-x^2}dx+{\int }_{-1}^{1}xdx-{\int }_{-1}^{1}dx+{\int }_1^2\sqrt{5-x^2}dx-{\int }_1^{2}xdx+{\int }_1^{2}dx}$

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

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

Required area= ${\displaystyle [\frac{5}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)-\frac{1}{2}]sq.units}$