Answer
$b = ln(3e^a-2)$
Work Step by Step
$A = \int_{0}^{a}e^x = e^a-e^0 = e^a-1$
$B = \int_{0}^{b}e^x = e^b-e^0 = e^b-1$
It is given that $B = 3A$:
$B = 3A$
$e^b-1 = 3(e^a-1)$
$e^b-1 = 3e^a-3$
$e^b = 3e^a-2$
$b = ln(3e^a-2)$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
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 402: 84
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