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: 75



Work Step by Step

$$A=\int^2_1\frac{4}{v^2}dv$$ $$A=\Big(-\frac{4}{v}\Big)\Big]^2_1$$ $$A=-\frac{4}{2}-\Big(-\frac{4}{1}\Big)=-2+4$$ $$A=2$$
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