University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 309: 32



Work Step by Step

Because $\sqrt 2\lt5\sqrt 2$, by Equation (2): $\int_{\sqrt 2}^{5\sqrt 2} r dr=\frac{(5\sqrt 2)^2}{2}-\frac{(\sqrt 2)^2}{2}=\frac{50}{2}-\frac{2}{2}=\frac{48}{2}=24$
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