<s>[INST] 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}$
 [/INST] To find the points of intersection between the curve $y = 2x - x^2$ and $y = -x$, equate the two expressions.
\begin{align*}
2x - x^2 & = -x\\
3x - x^2 & = 0\\
x(3 - x) & = 0
\end{align*}
which has solutions $x = 0$ and $x = 3$.  These are the limits of integration.
Since 
\begin{align*}
y & = 2x - x^2\\
  & = -x^2 + 2x\\
  & = -(x^2 - 2x)\\
  & = -(x^2 - 2x + 1) + 1\\
  & = -(x - 1)^2 + 1
\end{align*}
the graph of $y = 2x - x^2$ is a parabola with vertex $(1, 1)$ that opens downwards.  Hence, it is above the line $y = -x$ in the interval $(0, 3)$, as shown in the figure below.

We want to find the area of the shaded region.  Since it lies below the parabola and above the line, its area is 
$$\int_{0}^{3} [2x - x^2 - (-x)]~\textrm{d}x = \int_{0}^{3} (3x - x^2)~\textrm{d}x = \frac{9}{2}$$
</s>