Find by integration the area of the region bounded by the curve `y=2x-x^2` and the `x`-axis.

Explanation

The given curve is `y=2x-x^2`
Now `y=2x -x^2 rArr (x^2-2x + 1)=(-y+1)`
`rArr (x-1)^2=-1( y-1)`
`rArr X^2=-y`

where `(x-1)=X and (y-1) =y`
Clearly , `x^2 =-y` is a downward parabola with its vertex at `(X =0, Y=0 ).`
Now , `X=0 , Y=0 rArr x-1 =0 and y-1 =0`
`rArr x=1 and y=1`
Thus the vertex of the parabola is `A(1,1)`
Also `y=0 rArr 2x -x^2 =0 rArr x(2-x)=0 rArr x=0 or x=2`
Thus, the curve the x-axist at `O (0,0) and B (2,0)`
The rough sketch of the curve can now be drawn as shown in the given figure
`therefore` required area`= underset(0)overset(2)int y dx`
`= underset(0)overset(2)int(2x -x^2 ) dx = [x^2 - (x^3)/(3)]_0 ^-2`
`= (4-8/3)=4/3` sq units
Hence the required area is `4/3` sq units .