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: 15



Work Step by Step

We have $$\int_{1/16}^{1} t^{1/4}dt=\frac{1}{5/4}t^{5/4}|_{1/16}^1=\frac{4}{5} -\frac{4}{5} \frac{1}{{16^{5/4}}}\\=\frac{4}{5}-\frac{1}{{40}}=\frac{31}{40}.$$
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