Answer
$\int_{-1}^{0}$ $3$($x^{3}$ $-$ $x$) $dx$ $+$ $\int_{0}^{1}$ $-$$3$($x^{3}$ $-$ $x$) $dx$
Work Step by Step
Because $y_{1}$ is the uppermost function in the left region, and the lowermost in the right region, we must find two integrals and take their sum to find the total area.
For the interval $-1$ to $0$, $y_{2}$ $\lt$ $y_{1}$ .
The integral for this region is
$\int_{-1}^{0}$ ($3$($x^{3}$ $-$ $x$) $-$ $0$) $dx$
For the interval $0$ to $1$, $y_{2}$ $\gt$ $y_{1}$ .
The integral for this region is
$\int_{0}^{1}$ ($0$ $-$ $3$($x^{3}$ $-$ $x$)) $dx$
To obtain the total area, add the two integrals:
$\int_{-1}^{0}$ ($3$($x^{3}$ $-$ $x$) $-$ $0$) $dx$ $+$ $\int_{0}^{1}$ ($0$ $-$ $3$($x^{3}$ $-$ $x$)) $dx$
The zeroes can be eliminated to simplify to:
$\int_{-1}^{0}$ $3$($x^{3}$ $-$ $x$) $dx$ $+$ $\int_{0}^{1}$ $-$$3$($x^{3}$ $-$ $x$) $dx$
