# What is f(x) = int x^3-x if f(2)=4 ?

Jan 12, 2016

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

#### Explanation:

For solving the problem first integrate the indefinite integral.

$f \left(x\right) = \int \quad {x}^{3} - x \mathrm{dx}$

Use rule $\int \quad {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$

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

We are given $f \left(2\right) = 4$
We are going to find $C$ using the given value.

$f \left(2\right) = {\left(2\right)}^{4} / 4 - {\left(2\right)}^{2} / 2 + C$
$4 = \frac{16}{4} - \frac{4}{2} + C$
$4 = 4 - 2 + C$
4=2+C C=2#

Therefore,

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

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

#### Explanation:

Given: $f \left(x\right) = \int \left({x}^{3} - x\right)$$\mathrm{dx}$ and $f \left(2\right) = 4$

$f \left(x\right) = \int \left({x}^{3} - x\right)$$\mathrm{dx}$ = ${x}^{4} / 4 - {x}^{2} / 2 + C$

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

$f \left(2\right) = 4 = {2}^{4} / 4 - {2}^{2} / 2 + C$

and $C = 2$

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