Answer
$\displaystyle \frac{9}{2}$
Work Step by Step
Graph both functions in the same window (see below).
On the interval $x\in[0,3]$, the graph of $y= x$ is above the graph of $y=x^{2}-2x.$
The area between two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$.
Thus,
$A=\displaystyle \int_{0}^{3}[x-(x^{2}-2x)]dx=\int_{0}^{3}(3x-x^{2})dx$
$=3[\displaystyle \frac{x^{2}}{2}]_{0}^{3}-[\frac{x^{3}}{3}]_{0}^{3}$
$=\displaystyle \frac{3}{2}(9-0)-\frac{1}{3}(27-0)$
$=\displaystyle \frac{27}{2}-9$
$=\displaystyle \frac{27-18}{2}$
$=\displaystyle \frac{9}{2}$