Find the integral of f'(x) = 3x^2 - 6x + 3 given that f(0) = -4

Given the derivative, f;(x) = 3x^2 - 6x +
3  

Also, given f(0) = -4.

We need to
determine the function f(x).

We know
that:

f(x) = integral f'(x) :

Then, let
us integrate f'(x)"

==> f(x) = intg
f'(x)

                = intg ( 3x^2 - 6x + 3)
dx

                = 3x^3/3 - 6x^2/ 2 + 3x +
C

==> f(x) = x^3 - 3x^2 + 3x +
C

But given that f(0) = -4

==>
f(0) = 0 - 3*0 + 3*0 + C = -4

==> C =
-4

==> f(x) = x^3 - 3x^2 + 3x -
4