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: 41

Answer

$= \frac{15}{\ln 2} $

Work Step by Step

$\int^{4}_0 2^{s} ds$ $= \frac{2^{s}}{\ln 2} |^{4}_0$ $= [\frac{2^{4}}{\ln 2} ] - [\frac{2^{0}}{\ln 2} ]$ $= [\frac{16}{\ln 2} ] - [\frac{1}{\ln 2} ]$ $= \frac{16-1}{\ln 2} $ $= \frac{15}{\ln 2} $
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