Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 29



Work Step by Step

$\int^{ln2}_{0}e^{3x}dx$ =$\frac{1}{3}e^{3x} |^{ln2}_0$ =$\frac{1}{3}e^{3ln2}-\frac{1}{3}e^{ln8}-\frac{1}{3}$ =$\frac{8}{3}-\frac{1}{3}$ =$\frac{7}{3}$
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