# What is f(x) = int (3x-2)^2-5x+1 dx if f(2) = 1 ?

May 29, 2016

$f \left(x\right) = 3 {x}^{3} - \frac{17}{2} {x}^{2} + 5 x + 1$

#### Explanation:

The indefinite integral is

$\int \left({\left(3 x - 2\right)}^{2} - 5 x + 1\right) \setminus \mathrm{dx} = \int \left(9 {x}^{2} - 17 x + 5\right) \setminus \mathrm{dx}$

$= 3 {x}^{3} - \frac{17}{2} {x}^{2} + 5 x + C$

Therefore, $f \left(x\right) = 3 {x}^{3} - \frac{17}{2} {x}^{2} + 5 x + C$ for some $C$. Since $f \left(2\right) = 1$, we must choose the value of $C$ so that $1 = 3 \cdot {2}^{3} - \frac{17}{2} \cdot {2}^{2} + 5 \cdot 2 + C$. This implies that $1 = 24 - 34 + 10 + C$ so that $C = 1$.

Hence, the answer is $f \left(x\right) = 3 {x}^{3} - \frac{17}{2} {x}^{2} + 5 x + 1$.