Answer
$\frac{7}{3}$
Work Step by Step
$\int^{ln2}_{0}e^{3x}dx$
=$\frac{1}{3}e^{3x} |^{ln2}_0$
=$\frac{1}{3}e^{3ln2}-\frac{1}{3}e^{ln8}-\frac{1}{3}$
=$\frac{8}{3}-\frac{1}{3}$
=$\frac{7}{3}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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