The area of the region bounded by y=2x-x^2 and x-axis is :

The correct answer is Option (2) → $\frac{4}{3}$ sq.units

$y=2x-x^2-1+1$

$(y-1)=-(x-1)^2$

curve intersects x axis at y = 0

$⇒2x=x^2$

so $x=0,2$

Required area $=\int\limits_0^22x-x^2dx$

$=\left[x^2-\frac{x^3}{3}\right]_0^2$

$=\frac{4}{3}$ sq.units