Answer
The given improper integral
$$
\int_{-\infty}^{-27} x^{-5 / 3} d x=-\frac{1}{6}.
$$
is convergent
Work Step by Step
$$
\int_{-\infty}^{-27} x^{-5 / 3} d x
$$
Observe that the given integral is improper
$$
\begin{split}
\int_{-\infty}^{-27} x^{-5 / 3} d x&=\lim _{a \rightarrow-\infty} \int_{a}^{-27} x^{-5 / 3} d x\\
&=\left(\lim _{a \rightarrow-\infty}\left(\frac{x^{-2 / 3}}{-\frac{2}{3}}\right)\right|_{a} ^{-27}\\
&=\left.\lim _{a \rightarrow-\infty}\left(-\frac{3}{2} x^{-2 / 3}\right)\right|_{a} ^{-27}\\
&=\lim _{a \rightarrow-\infty}\left[-\frac{3}{2}(-27)^{-2 / 3}+\frac{3}{2}(a)^{-2 / 3}\right]\\
&=\lim _{a \rightarrow-\infty}\left(-\frac{1}{6}+\frac{3}{a^{2 / 3}}\right)\\
&=-\frac{1}{6}+\lim _{a \rightarrow-\infty}\left(\frac{3}{a^{2 / 3}}\right)\\
&=-\frac{1}{6}+0\\
&=-\frac{1}{6}.
\end{split}
$$
Thus the given improper integral
$$
\int_{-\infty}^{-27} x^{-5 / 3} d x=-\frac{1}{6}.
$$
