Answer
$$\dfrac{9}{2}$$
Work Step by Step
Simplify the improper integral $\int_{-1}^{8} x^{-1/3} \ dx$
Let us consider that $I=\int_{-1}^{8} x^{-1/3} \ dx=\int_{-1}^{0} x^{-1/3} \ dx+\int_{0}^{8} x^{-1/3} \ dx$
Integrate the integral by using limits.
$I=[\dfrac{x^{2/3}}{2/3}]_{-1}^{0}+[\dfrac{x^{2/3}}{2/3}]_{0}^{8}\\=(0-\dfrac{3}{2})+(6-0)\\=-\dfrac{3}{2}+6\\=\dfrac{9}{2}$
