Answer
$-5$
Work Step by Step
We have
$$\int_{8/27}^{1}(10t^{4/3}-8t^{1/3}) t^{-2} d t=\int_{8/27}^{1}(10t^{-2/3}-8t^{-5/3}) d t\\
=\frac{10t^{1/3}}{1/3}-\frac{8t^{-2/3}}{-2/3} |_{1/27}^{1}=30+12-30(2/3)-12(9/4)\\
= -5.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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