Answer
$8$
Work Step by Step
The area between two graphs over [a,b], where one graph is above the other, is given with
$\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
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Graphing, due to symmetry, the total area is 2$\times$(area right of the y-axis).
For $x\in[0,1],\ \ y_{above}=x^{4}-4x^{2}+4,\ \ y_{below}=x^{2}$
$A_{1}=\displaystyle \int_{0}^{1}(x^{4}-5x^{2}+4)dx=\left[\frac{x^{5}}{5}-\frac{5x^{3}}{3}+4x \right]_{0}^{1}=\frac{1}{5}-\frac{5}{3}+4=\frac{38}{15}$
For $x\in[1,2],\ \ y_{above}=x^{2},\ \ y_{below}=x^{4}-4x^{2}+4$
$A_{2}=-\displaystyle \int_{1}^{2}(x^{4}-5x^{2}+4)dx=\\\\\displaystyle -\left[\frac{x^{5}}{5}-\frac{5x^{3}}{3}+4x \right]_{1}^{2}=-[ \frac{32-1}{5}-\frac{5(8-1)}{3}+4(2-1) ]=\frac{22}{15}$
Total area = $2(A_{1}+A_{2})=2(\displaystyle \frac{38}{15}+\frac{22}{15})=8$
