Answer
$\displaystyle\int\limits_1^8x^{-2/3}dx=3$
Work Step by Step
$\displaystyle\int\limits_1^8x^{-2/3}dx$
Evaluate the integral and then apply the second part of the fundamental theorem of calculus to get the answer:
$\displaystyle\int\limits_1^8x^{-2/3}dx=\dfrac{1}{1-\frac{2}{3}}x^{1-\frac{2}{3}}\Big|_1^8=\dfrac{1}{\frac{1}{3}}x^{1/3}\Big|_1^8=3x^{1/3}\Big|_1^8=3\sqrt[3] x\Big|_1^8=...$
$...=3[\sqrt[3] 8-\sqrt[3] 1]=3(2-1)=3$