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: 76



Work Step by Step

$$A=\int^{27}_1x^{-4/3}dx$$ $$A=\Big(\frac{x^{-1/3}}{-\frac{1}{3}}\Big)\Big]^{27}_1=\Big(-\frac{3}{x^{1/3}}\Big)\Big]^{27}_1$$ $$A=-\frac{3}{(27)^{1/3}}-\Big(-\frac{3}{(1)^{1/3}}\Big)$$ $$A=-\frac{3}{3}+\frac{3}{1}=-1+3$$ $$A=2$$
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