Answer
$\frac{1}{6}$
Work Step by Step
Step 1. See figure. Rewrite the functions as $y=x^3$ and $y=x^5$. Find the intersections between the two functions. Letting $y=x^3=x^5$, we have $x=0,\pm1$. Thus, the intersections are $(-1,-1), (0,0), (1,1)$.
Step 2. The enclosed area between the two functions can be written as
$A=\int_{-1}^0 (x^5-x^3)dx + \int_0^1 (x^3-x^5)dx =\frac{1}{6}x^6|_{-1}^0-\frac{1}{4}x^4|_{-1}^0 + \frac{1}{4}x^4|_{0}^1 - \frac{1}{6}x^6|_{0}^1=-\frac{1}{6}(-1)^6+\frac{1}{4}(-1)^4 +\frac{1}{4}(1)^4-\frac{1}{6}(1)^6=\frac{1}{6}$
