## Calculus: Early Transcendentals 8th Edition

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