Answer
$$\frac{31}{40}.$$
Work Step by Step
We have
$$\int_{1/16}^{1} t^{1/4}dt=\frac{1}{5/4}t^{5/4}|_{1/16}^1=\frac{4}{5} -\frac{4}{5} \frac{1}{{16^{5/4}}}\\=\frac{4}{5}-\frac{1}{{40}}=\frac{31}{40}.$$
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: 15
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