Answer
$\frac{1}{2}$
Work Step by Step
Step 1. See figure. Find the intersections between the two functions. Letting $x=y^3=y$, we have $y=0,\pm1$ and $x=0,\pm1$. Thus the intersections are $(-1,-1), (0,0), (1,1)$
Step 2. The enclosed area between the two functions can be written as (integrate over $dy$)
$A=\int_{-1}^0 (y^3-y)dy + \int_0^1 (y-y^3)dy =\frac{1}{4}y^4|_{-1}^0-\frac{1}{2}y^2|_{-1}^0 + \frac{1}{2}y^2|_{0}^1 - \frac{1}{4}y^4|_{0}^1=-\frac{1}{4}(-1)^4+\frac{1}{2}(-1)^2 +\frac{1}{2}(1)^2-\frac{1}{4}(1)^4=\frac{1}{2}$
