Answer
$\displaystyle \frac{48}{5}$
Work Step by Step
Graph both functions in the same window (see below).
On the interval $x\in[0,2]$, the graph of $y= 8x$ is above the graph of $y=x^{4}.$
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,
$A=\displaystyle \int_{0}^{2}[8x-(x^{4})]dx=$
$=8[\displaystyle \frac{x^{2}}{2}]_{0}^{2}-[\frac{x^{5}}{5}]_{0}^{2}$
$=4(4-0)-\displaystyle \frac{1}{5}(32-0)$
$=16-\displaystyle \frac{32}{5}$
$=\displaystyle \frac{80-32}{5}$
$=\displaystyle \frac{48}{5}$
