Answer
$\displaystyle \frac{27}{4}$
Work Step by Step
Expressing the $y$'s as functions of $x$,
$y=f(x)=x^{3},$
$y=g(x)=3x^{2}-4.$
Graphing the functions, we find the intersection points at $x=-1$ and $x=2$.
Over the interval $x\in[-1,2],\ f(x)$ is above $g(x)$, so
$A=\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx=\int_{-1}^{2}[x^{3}-(3x^{2}-4)]dx$.
$=\displaystyle \int_{-1}^{2}[x^{3}-3x^{2}+4]dx$.
$=\displaystyle \left[\frac{x^{4}}{4}-x^{3}+4x \right]_{-1}^{2}$
$=\displaystyle \frac{16-1}{4}-(8+1)+4(2+1)$
$=\displaystyle \frac{15}{4}-9+12$
=$\displaystyle \frac{27}{4}$
