Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$ $$
y =\ 2\ {x} - {x}^2
$$
$$
y =\ -{x}
$$
According to me , the area
$$
\int_{0}^{2}{2x\ -\ { x} ^2}\, dx \ + \int_{2}^{3}{\ {x} ^2\  -\ 2{x} }\, dx \\
$$
Which gives the area $ \frac{8}{3}$
But the answer is $ \frac{9}{2}$

You should go around the path in clockwise fashion:
$$
\int_0^3(2x-x^2)\,dx+\int_3^0(-x)\,dx=
\int_0^3(2x-x^2+x)\,dx=\Bigl[\frac{3}{2}x^2-\frac{1}{3}x^3\Bigr]_0^3=\frac{27}{2}-9=\frac{9}{2}
$$
