Given equation of curve is $y=2x-x^{2}$
$\Rightarrow x^{2}-2 x=-y$
$\Rightarrow x^{2}-2 x+1=-y+1 $
$\Rightarrow (x-1)^{2}=-(y-1)$
This is the equation of parabola having vertex (1,1) and open downward.
The parabola intersect the $X$ -axis, put $y=0,$ we get
$0 =2 x-x^{2} $
$\Rightarrow x (2-x) =0$
$\Rightarrow x=0,2$
$\therefore $ Area of bounded region between the curve and $X$ -axis
$=\int\limits_{0}^{2} ydx$
$=\int\limits_{0}^{2}\left(2 x-x^{2}\right) d x=\left[\frac{2 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}$
$=\left[4-\frac{8}{3}-0-0\right]=\frac{4}{3}$ sq units.
