# What is f(x) = int (3x-1)^2-2x+1 dx if f(2) = 1 ?

Jan 21, 2016

$f \left(x\right) = 3 {x}^{3} - 4 {x}^{2} + 2 x - 11$

#### Explanation:

Now, the given function is apparently an integral of another function.
That is $f \left(x\right) = \int \left({\left(3 x - 1\right)}^{2} - 2 x + 1\right) \mathrm{dx}$
Before integrating, let's expand and simplify the terms shall we?

${\left(3 x - 1\right)}^{2} - 2 x + 1 = 9 {x}^{2} + 1 - 6 x - 2 x + 1 = 9 {x}^{2} - 8 x + 2$
So, $\int \left(9 {x}^{2} - 8 x + 2\right) \mathrm{dx} = 9 {x}^{3} / 3 - 8 {x}^{2} / 2 + 2 x + c$
$\setminus \implies f \left(x\right) = 3 {x}^{3} - 4 {x}^{2} + 2 x + c$

Given that at $x = 2 \text{ } f \left(x\right) = 1$
So, $3 \cdot {2}^{3} - 4 \cdot {2}^{2} + 2 \cdot 2 + c = 1 \setminus \implies 3 \cdot 8 - 4 \cdot 4 + 4 + c = 1$
$\therefore 24 - 16 + 4 + c = 1 \setminus \implies c = - 11$
Substituting for $c$ we get the answer I have written above.