Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 72


The mass of the rod is $20~grams$

Work Step by Step

Let $M(x)$ be the mass of the rod of length $x$ $M'(x) = \rho(x) = \frac{1}{\sqrt{x}} = x^{-1/2}$ We can find an expression for $M(x)$: $M(x) = 2~\sqrt{x}+C$ Since $M(0) = 0,$ then $C = 0$ $M(x) = 2~\sqrt{x}$ We can find $M(100)$: $M(x) = 2~\sqrt{100} = 20~grams$ The mass of the rod is $20~grams$
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