University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 29

Answer

$\frac{7}{3}$

Work Step by Step

$\int_{0}^{ln2} (e^{3x}) dx=[\frac{1}{3}\cdot e^{3x}]_{0}^{ln2}=[\frac{1}{3}\cdot e^{3\cdot ln2}]-[\frac{1}{3}\cdot e^{3\cdot 0}]=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}$
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