The improper integral $\int_{-8}^1\frac{dx }{x^{1/3}}$ is evaluated as
$\lim_{a \to 0^-} \int _{-8}^a x^{-1/3} dx + \lim_{a \to 0^+} \int _{a}^1 x^{-1/3} dx=-9/2$
the graph of the function for negative x is below the x axis, so why are we not considering the first of the two integrals with a negative sign ?
$\lim_{a \to 0^-} \color{red}{-(\int _{-8}^a x^{-1/3} dx)} + \lim_{a \to 0^+} \int _{a}^1 x^{-1/3} dx=15/2$
I think you're mixing two things: calculating the integral and finding an area bounded by the graph and the $x$-axis on a certain interval.
If you integrate $\sin x$ on $[0,2\pi]$, you'll find zero: $$\int_0^{2\pi} \sin x \,\mbox{d}x = [-\cos x]_0^{2\pi} = 0$$ because the area under the graph and above the $x$-axis on $[0,\pi]$ is the same as the area above the graph and under the $x$-axis on $[\pi,2\pi]$. As a consequence of how integrals are defined, that area below the $x$-axis gets a minus-sign so they cancel and give $0$ as a result.
Now if you want to find the area bounded by the graph and the $x$-axis, you would manually add an extra minus-sign on the interval(s) where the graph is below the $x$-axis so all area is added up (of course in this case, you could just double the value on $[0,\pi]$ because of symmetry). In fact, you integrate $|\sin x|$ but in manual calculations, that would result in splitting and adding the minus: $$\int_0^{2\pi} |\sin x| \,\mbox{d}x = \int_0^{\pi} \sin x \,\mbox{d}x \color{red}{-} \int_\pi^{2\pi} \sin x \,\mbox{d}x = \ldots = 1\color{red}{-} (-1) = 2$$
Back to your problem: if you simply need (or want) to calculate the integral, you get: $$\int_{-8}^1\frac{1}{x^{1/3}}\,\mbox{d}x = \ldots = -\frac{9}{2}$$ If you want to find the area, you would integrate $|x^{-1/3}|$ and that would require adding the minus-sign on the interval $[-8,0)$ because the function is negative there: $$\int_{-8}^1\left|\frac{1}{x^{1/3}}\right|\,\mbox{d}x = \color{red}{-}\int_{-8}^0\frac{1}{x^{1/3}}\,\mbox{d}x+ \int_{0}^1\frac{1}{x^{1/3}}\,\mbox{d}x = \ldots = \frac{15}{2}$$
Note: in this case of an improper integral, the region 'bounded' by the graph of the function and the $x$-axis isn't bounded, since it stretches to infinity near the $y$-axis. However, if the improper integral converges, we still assign that finite value as area to this unbounded region.