Answer
$$\frac{4}{3}$$
Work Step by Step
Since
\begin{aligned}
y_{1}(x) &=y_{2}(x) \\
x^{3}-x &=1-x^{2} \\
x^{3}+x^{2}-x-1 &=0 \\
(x-1)(x+1)^{2} &=0 \\
x=\pm 1
\end{aligned}
Then
\begin{aligned}
A &=\int_{-1}^{1}\left(y_{1}-y_{2}\right) d x \\
&=\int_{-1}^{1}\left(\left(1-x^{2}\right)-x\left(x^{2}-1\right)\right) d x \\
&=\int_{-1}^{1}\left(\left(1-x^{2}\right)-x\left(x^{2}-1\right)\right) d x \\
&=\left(-x^{3}-x^{2}+x+1\right) d x \\
&=\left.\left(-\frac{x^{4}}{4}-\frac{x^{3}}{3}+\frac{x^{2}}{2}+x\right)\right|_{-1} ^{1} \\
&=\left(-\frac{1}{4}-\frac{1}{3}+\frac{1}{2}+1\right)-\left(-\frac{1}{4}-\frac{-1}{3}+\frac{1}{2}-1\right) \\
&=\frac{4}{3}
\end{aligned}
