Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 54



Work Step by Step

We have $$ \int_2^3 (7y-5)^{1/2}dy=\frac{1}{7}\int_2^3 7(7y-5)^{1/2}dy\\ =\frac{1}{7} \frac{1}{3/2}(7y-5)^{3/2}|_2^3\\ =\frac{2}{21}(16^{3/2}-9^{3/2})\\ =\frac{2}{21}(64-27) =\frac{74}{21}. $$
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