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: 57



Work Step by Step

$$A=\int^4_{0}\frac{2t}{t^2-25}dt$$ We set $a=t^2-25$, which means $$da=2tdt$$ For $t=0$, $a=-25$ and for $t=4$, $a=-9$ Therefore, $$A=\int^{-9}_{-25} \frac{1}{a}da$$ $$A=\Big(\ln|a|\Big)\Big]^{-9}_{-25}$$ $$A=\ln9-\ln25=\ln\frac{9}{25}$$
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