We are given f'(x) = x^2 - 5x - 3 and f(0) = -2

To find f(x) we have to integrate f'(x)

Int [ f'(x)] = Int [ x^2 - 5x - 3]

=> f(x) = x^3 / 3 - (5/2)x^2 - 3x + C

f(0) = 0 + 0 + 0 + C = -2

=> C = -2

f(x) = x^3/ 3 - (5/2)x^2 - 3x - 2

f'(x) = x^2 -5x -3

We need to find f(x).

We know that f(x) = integral of f'(x).

==> f(x) = Int ( x^2 -5x -3) dx

             = Int (x^2) dx - Int (5x) dx - Int 3 dx

              = x^3/3 - 5x^2/2 - 3x + C

==> f(x)= (1/3)x^3 -(5/2)x^2 - 3x + C

But we are given that f(0) = -2

==> f(0) = 0 - 0 - 0 + C = -2

==> C = -2

==> f(x) = (1/3)x^3 - (5/2)x^2 - 3x -2