Answer
$=\displaystyle \frac{5}{6}$
Work Step by Step
On the interval $x\in[0,1]$, the graph of $y= x$ is above the graph of $y=\displaystyle \frac{x^{2}}{4}$
On the interval $x\in[1,2]$, the graph of $y= 1$ is above the graph of $y=\displaystyle \frac{x^{2}}{4}$
The area between the 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,
$A=\displaystyle \int_{0}^{1}[x-\frac{x^{2}}{4}]dx+\int_{1}^{2}[1-\frac{x^{2}}{4}]dx$
$=[\displaystyle \frac{x^{2}}{2}]_{0}^{1}-\frac{1}{4}[\frac{x^{3}}{3}]_{0}^{1}+[x]_{1}^{2}-\frac{1}{4}[\frac{x^{3}}{3}]_{1}^{2}$
$=(\displaystyle \frac{1}{2}-0)-\frac{1}{12}(1-0)+(2-1)-\frac{1}{12}(2^{3}-1)$
$=\displaystyle \frac{1}{2}-\frac{1}{12}+1-\frac{7}{12}$
$=\displaystyle \frac{6-1-7+12}{12}$
$=\displaystyle \frac{10}{12}$
$=\displaystyle \frac{5}{6}$
