Answer
$\frac{74}{21}.$
Work Step by Step
We have
$$
\int_2^3 (7y-5)^{1/2}dy=\frac{1}{7}\int_2^3 7(7y-5)^{1/2}dy\\
=\frac{1}{7} \frac{1}{3/2}(7y-5)^{3/2}|_2^3\\
=\frac{2}{21}(16^{3/2}-9^{3/2})\\
=\frac{2}{21}(64-27)
=\frac{74}{21}.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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