# How do you find the antiderivative of int (2x^3-1)(x^4-2x)^6dx from [-1,1]?

##### 1 Answer
Nov 9, 2016

${\int}_{- 1}^{1} \left(2 {x}^{3} - 1\right) {\left({x}^{4} - 2 x\right)}^{6} \mathrm{dx} = \frac{- 1094}{7}$

#### Explanation:

${\int}_{- 1}^{1} \left(2 {x}^{3} - 1\right) {\left({x}^{4} - 2 x\right)}^{6} \mathrm{dx}$

We will apply the substitution $u = {x}^{4} - 2 x$. Differentiating this reveals that $\mathrm{du} = \left(4 {x}^{3} - 2\right) \mathrm{dx}$. Note that the $\left(2 {x}^{3} - 1\right)$ term is exactly half this, so we can modify the integral as follows:

$= \frac{1}{2} {\int}_{- 1}^{1} {\left({x}^{4} - 2 x\right)}^{6} \left(4 {x}^{3} - 2\right) \mathrm{dx}$

Now we can substitute in our $u$ and $\mathrm{du}$ values. However, we also should remember to change our bounds! Take $- 1$ and $1$ and plug them into our $u = {x}^{4} - 2 x$ term. That is, the bound of $- 1$ becomes ${\left(- 1\right)}^{4} - 2 \left(- 1\right) = 3$ and the bound of $1$ becomes ${1}^{4} - 2 \left(1\right) = - 1$. Thus:

$= \frac{1}{2} {\int}_{3}^{- 1} {u}^{6} \mathrm{du} = \frac{1}{2} {\left[{u}^{7} / 7\right]}_{3}^{- 1} = \frac{1}{2} \left[{\left(- 1\right)}^{7} / 7 - {3}^{7} / 7\right]$

$= \frac{1}{2} \left[\frac{- 2188}{7}\right] = \frac{- 1094}{7}$