Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 24



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: $\displaystyle\int\limits_1^8x^{-2/3}dx=\dfrac{1}{1-\frac{2}{3}}x^{1-\frac{2}{3}}\Big|_1^8=\dfrac{1}{\frac{1}{3}}x^{1/3}\Big|_1^8=3x^{1/3}\Big|_1^8=3\sqrt[3] x\Big|_1^8=...$ $...=3[\sqrt[3] 8-\sqrt[3] 1]=3(2-1)=3$
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