Explanation:
First, find the intersections.
4x = x³
0 = x³ − 4x
0 = x (x² − 4)
0 = x (x + 2) (x − 2)
x = -2, 0, 2
So we will consider two intervals: -2 ≤ x ≤ 0, and 0 ≤ x ≤ 2.
Pick a point within each interval and evaluate both functions at that point. Compare to see which function is greater.
At x = -1, y₁ = 4x = -4, and y₂ = x³ = -1. So y₂ > y₁.
At x = 1, y₁ = 4x = 4, and y₂ = x³ = 1. So y₁ > y₂.
So the integral would be:
∫₋₂⁰ (x³ − 4x) dx + ∫₀² (4x − x³) dx