# How do you find antiderivative of (1-x)^2?

May 3, 2018

${\left(x - 1\right)}^{3} / 3 + c$

#### Explanation:

$\int {\left(1 - x\right)}^{2} \mathrm{dx} =$

Substitute $1 - x = u$

$- \mathrm{dx} = \mathrm{du}$
$\mathrm{dx} = - \mathrm{du}$

• $\int {u}^{2} \left(- \mathrm{du}\right)$ $=$

$- \int {u}^{2} \mathrm{du}$ $=$

$- \int \left({u}^{3} / 3\right) ' \mathrm{du}$ $=$

$- {u}^{3} / 3 + c$ $=$

${\left(x - 1\right)}^{3} / 3 + c$

, $c$$\in$$\mathbb{R}$