Answer
$$\int^2_1\frac{4}{v^2}dv=2$$
Work Step by Step
$$A=\int^2_1\frac{4}{v^2}dv$$ $$A=\Big(-\frac{4}{v}\Big)\Big]^2_1$$ $$A=-\frac{4}{2}-\Big(-\frac{4}{1}\Big)=-2+4$$ $$A=2$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Practice Exercises - Page 343: 75
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