To find the definite integral of the function F(x) over the interval [-4, 0], we need to find the antiderivative of f(x) first and then evaluate it at the upper and lower bounds of the interval.

First, let's find the antiderivative of f(x):
F(x) = x^3 + 3x^2 - 9x

Integrating term by term, we get:
∫F(x) dx = ∫(x^3 + 3x^2 - 9x) dx
= (1/4)x^4 + (1)x^3 - (9/2)x^2 + C

Now, we will evaluate the definite integral over the interval [-4, 0]:
∫F(x) dx from -4 to 0 = [ (1/4)(0)^4 + (1)(0)^3 - (9/2)(0)^2 ] - [ (1/4)(-4)^4 + (1)(-4)^3 - (9/2)(-4)^2 ]
= [ 0 + 0 - 0 ] - [ (1/4)(256) + (-64) - 72 ]
= 0 - [ 64 + (-64) - 72 ]
= -64 + 64 + 72
= 72

Therefore, the definite integral of the function F(x) over the interval [-4, 0] is 72.