# What is f(x) if f'(x)=6x^2-2x+1 ?

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### 2 Answers

f'(x) = 6x^2-2x+1. To find f(x).

f(x) is the integral of f'x) dx or f(x) is the integral (6x^2-2x+1) dx.

f(x) = Int (6x^2-2x+1) dx.

f(x) = Int (6x^2)dx - Int 2xdx +Integral dx, As Inf (f(x) +or- g(x))dx = Int f(x)d +or-Intg(x)dx.

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

f(x) = 6 (1/2+1)x^(2+1) - 2(1/1+1) x^2 + x +C , as In x^n dx = (1/n+1)x^(n+1).

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

We know that F'(x) = f(x), where Int f'(x) dx = f(x).

We'll apply the indefinite integral to the expression of f(x):

Int f'(x) dx = Int (6x^2-2x+1)dx

We'll apply the additive property of indefinite integrals:

Int (6x^2-2x+1)dx = Int 6x^2dx - Int 2xdx + Int dx

Int 6x^2dx = 6x^3/3 + C

We'll simplify and we'll get:

Int 6x^2dx = 2x^3 + C (1)

Int 2xdx = 2x^2/2 + C

We'll simplify and we'll get:

Int 2xdx = x^2 + C (2)

Int dx = x + C (3)

We'll add (1),(2),(3):

f(x) = (1)+(2)+(3)

**f(x) = 2x^3 + x^2 + x + C**

Note: C+C+C = C (family of constants)