Answer
$\displaystyle \frac{5}{6}$
Work Step by Step
On the interval $x\in[0,1]$, the graph of $y= x^{2}$ is above the graph of $y=0$
On the interval $x\in[1,2]$, the graph of $y= -x+2$ is above the graph of $y=0$
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,
$A=\displaystyle \int_{0}^{1}[x^{2}-0]dx+\int_{1}^{2}(-x+2-0]dx=$
$=\displaystyle \int_{0}^{1}x^{2}dx-\int_{1}^{2}xdx+\int_{1}^{2}2dx$
$=[\displaystyle \frac{x^{3}}{3}]_{0}^{1}-[\frac{x^{2}}{2}]_{1}^{2}+2[x]_{1}^{2}$
$=(\displaystyle \frac{1}{3}-0)-\frac{2^{2}-1}{2}+2(2-1)$
$=\displaystyle \frac{1}{3}-\frac{3}{2}+2$
$=\displaystyle \frac{2-9+12}{6}$
$=\displaystyle \frac{5}{6}$
