Calculus: Area Between Curve and x-axis

Find the total area between the curve, y = x^(1/3) - x, and the x-axis for -1<= x <= 15?

When −1 < x < 0 ----> y < 0
When 0 < x < 1 ----> y > 0
When 1 < x < 15 -----> y < 0

A = ∫₋₁¹⁵ |x^(1/3) − x| dx
A = ∫₋₁⁰ (x − x^(1/3)) dx + ∫₀¹ (x^(1/3) − x) dx + ∫₁¹⁵ (x − x^(1/3)) dx
A = (1/2 x² − 3/4 x^(4/3)) |₋₁⁰ + (3/4 x^(4/3) − 1/2 x²) |₀¹ + (1/2 x² − 3/4 x^(4/3)) |₁¹⁵
A = 0 − (1/2 − 3/4) + (3/4 − 1/2) − 0 + (225/2 − 45∛15/4) − (1/2 − 3/4)
A = 1/4 + 1/4 + 225/2 + 1/4 − 45∛15/4
A = (453 − 45∛15)/4
A ≈ 85.505114164

For teaching purposes, I'm going to call the first curve f(x) and the second curve g(x).
Find the total area between f(x) = x^(1/3) - x and g(x) = 0, for -1 <= x <= 15.

Our first step is NOT going to evaluate the area using -1 and 15 as bounds; we first need to see where these curves cross.
Determine the points of the two curves' intersection by equating them to each other and then solving for x.

f(x) = g(x)
x^(1/3) - x = 0
x^(1/3) (1 - x^(2/3)) = 0
Which implies
x^(1/3) = 0; therefore x = 0.
1 - x^(2/3) = 0
-x^(2/3) = -1
x^(2/3) = 1
x^(1/3) = +/- 1
x = +/- 1

So x = { -1, 0, 1 }

The fact that the two curves cross at points within the interval -1 <= x <= 15 indicates we need to add many different areas.
What you have to do is determine, on the interval a to b, which of f(x) or g(x) is greater.
The reason why we want to do this is because
Area = Integral (a to b, [higher curve] - [lower curve] dx )

The five areas we are concerned with are:
1) -1 <= x <= 0
2) 0 <= x <= 1
3) 1 <= x <= 15

Part 1: Find the area between -1 and 0.
Which curve is higher? Test x = -1/8.
f(-1/8) = (-1/8)^(1/3) = -1/2
g(-1/8) = 0
Therefore, g(x) is higher, and the formula for the first part is
A[first] = Integral (-1 to 0, [g(x) - f(x)] dx )

Part 2: Find the area between 0 and 1.