Evaluate the integral or show that it is divergent?

Jan 4, 2018

The improper integral converges and equals $\frac{32}{3}$.

Explanation:

First note that

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

Next, note that the "impropriety" in the definite integral occurs at $x = - 2$ because the function has a vertical asymptote there. Hence,

${\int}_{- 2}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = {\lim}_{a \to - 2 +} {\int}_{a}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right)$

assuming this limit exists, where the positive sign to the right of $- 2$ in the limit indicates that $a$ is approaching $- 2$ from the right.

${\int}_{a}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = \frac{4}{3} {\left(x + 2\right)}^{\frac{3}{4}} {|}_{a}^{14}$

$= \frac{4}{3} \cdot {16}^{\frac{3}{4}} - \frac{4}{3} {\left(a + 2\right)}^{\frac{3}{4}} = \frac{32}{3} - \frac{4}{3} {\left(a + 2\right)}^{\frac{3}{4}}$.

Since ${\lim}_{a \to - 2 +} {\left(a + 2\right)}^{\frac{3}{4}} = {0}^{\frac{3}{4}} = 0$, it follows that the improper integral converges to $\frac{32}{3}$.

We therefore write

${\int}_{- 2}^{14} \frac{\mathrm{dx}}{x + 2} ^ \left(\frac{1}{4}\right) = \frac{32}{3}$.