Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 417: 73

Answer

The total mass of the rod is $~\frac{140}{3}~kg$

Work Step by Step

We can find the total mass of the rod: $\int_{0}^{4}\rho(x)~dx$ $=\int_{0}^{4}(9+2\sqrt{x})~dx$ $=(9x+\frac{4x^{3/2}}{3})\vert_{0}^{4}$ $=[9(4)+\frac{4(4)^{3/2}}{3}]+[9(0)+\frac{4(0)^{3/2}}{3}]$ $=(36+\frac{32}{3})+(0)$ $=(\frac{108}{3}+\frac{32}{3})$ $= \frac{140}{3}~kg$ The total mass of the rod is $~\frac{140}{3}~kg$
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