Find f(x) if f'(x) = 2x^2 - 5x + 7 and f(0) = -3

Given f'(x) = 2x^2-5x-7 to find f(x), if  f(0) =
-3.

First we find the f(x).

Since f'(x)
is given, f(x) = Int f'(x) dx.

Therefore f(x) = Int (2x^2-5x+7)
dx.

f(x) = Int 2x^2dx - Int5xdx + Int 7
dx.

f(x) =  2 (1/3)x^3 - 5(1/2)x^2 +7x +Const , as Int Kx^n dx =
K(1/n+1)x^(n+1) , for all n except n = -1.

Therefore f(x) =
(2/3)x^3-(5/2)x^2+7x +C....(1)

Now we use the condition f(0) = -3 to
find C:

f(0) = (2/3)*0^3-(5/2)*0^2+7*0 +C =
-3.

Therefore C = -3.

So we rewrite 
f(x) in (1) replacing C by -3:

f(x) = (2/3)x^3-(5/2)x^2+7x
-3.