Correct answer is \(\frac{4}{3}sq.units\)
Explaination::
Given curve is \( y =2 x-x^{2}\) or \((x-1)^{2} =-(y-1)\)
The curve cut the x -axis at (0,0) and (2,0) .
\(\therefore \ Required\ area =\int_{0}^{2} y d x\)
\(=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2}\)
\(=4-\frac{8}{3}=\frac{4}{3} sq \ unit\)