Answer
$\dfrac{-9}{2}$
Work Step by Step
Consider $f(x)=\int_{-8}^{1} \dfrac{1}{x^{1/3}} dx$
or, $f(x)=\int_{-8}^{0} \dfrac{1}{x^{1/3}} dx +\int_{0}^{1} \dfrac{1}{x^{1/3}} dx$
Thus, $\lim\limits_{a \to 0^{-}}\int_{-8}^{0} \dfrac{1}{x^{1/3}} dx +\lim\limits_{a \to 0^{+}}\int_{0}^{1} \dfrac{1}{x^{1/3}} dx=\lim\limits_{a \to 0^{-}} [\dfrac{3}{2}a^{2/3}-6]+\lim\limits_{a \to 0^{+}} [\dfrac{3}{2}-\dfrac{3}{2}a^{2/3}]$
or, $=-6+\dfrac{3}{2}$
or, $= \dfrac{-9}{2}$
