Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 24



Work Step by Step

We have $$\int_{8/27}^{1}(10t^{4/3}-8t^{1/3}) t^{-2} d t=\int_{8/27}^{1}(10t^{-2/3}-8t^{-5/3}) d t\\ =\frac{10t^{1/3}}{1/3}-\frac{8t^{-2/3}}{-2/3} |_{1/27}^{1}=30+12-30(2/3)-12(9/4)\\ = -5.$$
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