# How do you test the improper integral int (2x-1)^3 dx from (-oo, oo) and evaluate if possible?

Mar 28, 2018

Diverges.

#### Explanation:

First, let's obtain the indefinite integral $\int {\left(2 x - 1\right)}^{3} \mathrm{dx}$ so as to make our work easier when evaluating the improper integral:

$u = 2 x - 1 , \mathrm{du} = 2 \mathrm{dx} , \frac{1}{2} \mathrm{du} = \mathrm{dx}$

$\frac{1}{2} \int {u}^{3} \mathrm{du} = \frac{1}{8} {u}^{4} = \frac{1}{8} {\left(2 x - 3\right)}^{4}$

No need to include a constant of integration, this was only calculated to keep it on the side for when it becomes needed for evaluating the improper integral.

So,

${\int}_{-} {\infty}^{\infty} {\left(2 x - 3\right)}^{3} \mathrm{dx} = {\int}_{-} {\infty}^{0} {\left(2 x - 3\right)}^{3} \mathrm{dx} + {\int}_{0}^{\infty} {\left(2 x - 3\right)}^{3} \mathrm{dx}$

This sort of splitting up is necessary when evaluating from $\left(- \infty , \infty\right) .$ It's often easier to split up at zero (unless zero is not in the domain of the integrand).

$= {\lim}_{t \to - \infty} {\int}_{t}^{0} {\left(2 x - 3\right)}^{3} \mathrm{dx} + {\lim}_{t \to \infty} {\int}_{0}^{t} \left(2 x - 3\right) \mathrm{dx}$

$= {\lim}_{t \to - \infty} \left(\frac{1}{8} {\left(2 x - 3\right)}^{4}\right) {|}_{t}^{0} + {\lim}_{t \to \infty} \left(\frac{1}{8} {\left(2 x - 3\right)}^{4}\right) {|}_{0}^{t}$

$= {\lim}_{t \to - \infty} \left(\frac{1}{8} {\left(- 3\right)}^{4} - \frac{1}{8} {\left(2 t - 3\right)}^{4}\right) + {\lim}_{t \to \infty} \left(\frac{1}{8} {\left(2 t - 3\right)}^{4} - \frac{1}{8} {\left(- 3\right)}^{4}\right) = - \frac{81}{8} - \infty + \ldots$

The first limit did not exist, it went to negative infinity. No need to even evaluate the second one, we already know the entire integral must diverge.