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

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### 3 Answers

We have to find the integral of f'(x) = 3x^2 - 6x + 3. We have the value of f(0) given as -4.

Now Int f'(x)

=> Int [ 3x^2 - 6x + 3]

=> Int [ 3x^2] - Int [ 6x] + Int 3

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

=> x^3 - 3x^2 + 3x + C

Now f(0) = -4

=> 0^3 - 3*0^2 + 3*0 + C = -4

=> C = -4

Therefore the integral is x^3 - 3x^2 + 3x - 4.

**The required value of f(x) is x^3 - 3x^2 + 3x - 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**

f'(x) = 3x^2-6x+3. To find f(0).

Since f'(x) = 3x^2-6x+3,

f(x) = Int f(x) dx.

f(x) = Int (3x^2-6x+3) dx.

f(x) = Int (3x^2)dx -Int (6x)dx+Int(3dx)

f(x) = (3/3)x^3-(6/2)x^2 +3x +C .

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

f(0) = -4 = 0+C. So C = -4.

So we rewrite f(x) with C = -4.

f(x) = x^3-3x^2+3x-4 is the required function.