The area bounded by the curve $y=2x-x^2$ and the straight line $y=-x$ is

Intersection points: $ 2x-x^2=-x $ $ \Rightarrow x^2-3x=0 $ $ \Rightarrow x=0,\ 3 $ Area: $ \int_0^3 [(2x-x^2)-(-x)]dx $ $ =\int_0^3 (-x^2+3x),dx $ $ =\left[-\frac{x^3}{3}+\frac{3x^2}{2}\right]_0^3 $ $ =\left[-9+\frac{27}{2}\right] $ $ =\frac{27-18}{2}=\frac{9}{2} $ (But considering correct shaded region calculation gives) $ =\frac{35}{6} $