Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 22

You need to determine first the points of intersection between curves y = x^3 and y = x , by solving the equation, such that:
x^3 = x => x^3 - x = 0
Factoring out x yields:
x(x^2 - 1) = 0 => x = 0 or x^2 - 1 = 0 => x = 1 and x = -1
Hence, the endpoints of integral are x = -1, x = 0, x = 1.
You need to decide what curve is greater than the other on the interval [-1,1]. You need to notice that x^3>x on the interval [-1,0], and x^3int_a^b (f(x) - g(x))dx, where f(x) > g(x) for x in [a,b]
int_(-1)^1 |x^3 - x|dx = int_(-1)^0 (x^3 - x) dx + int_0^1 (x - x^3) dx
int_(-1)^1 |x^3 - x|dx = int_(-1)^0 x^3 dx - int_(-1)^0 xdx + int_0^1 x dx - int_0^1 x^3 dx
int_(-1)^1 |x^3 - x|dx = x^4/4|_(-1)^0 - x^2/2_(-1)^0 + x^2/2|_0^1 - x^4/4|_0^1
int_(-1)^1 |x^3 - x|dx = 0 - 1/4 - 0 + 1/2 + 1/2 - 0 - 1/4 + 0
int_(-1)^1 |x^3 - x|dx = 1 - 1/2
int_(-1)^1 |x^3 - x|dx = 1/2
Hence, evaluating the area of the region enclosed by the given curves, yields int_(-1)^1 |x^3 - x|dx = 1/2.

The area of the region enclosed by the given curves is found between the red and orange curves, for x in [-1,1].