University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 55



Work Step by Step

$$A=\int^1_{-1}\frac{dx}{3x-4}$$ We set $a=3x-4$, which means $$da=3dx$$ $$dx=\frac{1}{3}da$$ For $x=-1$, $a=-7$ and for $x=1$, $a=-1$ Therefore, $$A=\frac{1}{3}\int^{-1}_{-7} \frac{1}{a}da$$ $$A=\frac{1}{3}\Big(\ln|a|\Big]^{-1}_{-7}\Big)$$ $$A=\frac{1}{3}(\ln1-\ln7)$$ $$A=-\frac{\ln7}{3}$$
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