Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 410: 59


(a) $-\frac{3}{2}~m$ (b) $\frac{41}{6}~m$

Work Step by Step

(a) We can find the displacement: $\int_{0}^{3}v(t)~dt$ $=\int_{0}^{3}(3t-5)~dt$ $=(\frac{3t^2}{2}-5t)~\vert_{0}^{3}$ $=[\frac{3~(3)^2}{2}-5(3)]- [\frac{3(0)^2}{2}-5(0)]$ $=(\frac{27}{2}-15)- (0)$ $= -\frac{3}{2}~m$ (b) Note that the function $~3t-5~$ is negative on the interval $0 \leq t \lt \frac{5}{3}$ We can find the distance traveled: $\int_{0}^{3}\vert v(t) \vert~dt$ $=\int_{0}^{3}\vert 3t-5 \vert~dt$ $=\int_{0}^{5/3}\vert 3t-5 \vert~dt+\int_{5/3}^{3}\vert 3t-5 \vert~dt$ $=\int_{0}^{5/3}(5-3t)~dt+\int_{5/3}^{3}(3t-5)~dt$ $=(5t-\frac{3t^2}{2})~\vert_{0}^{5/3}+(\frac{3t^2}{2}-5t)~\vert_{5/3}^{3}$ $=[5~(\frac{5}{3})-\frac{3~(\frac{5}{3})^2}{2}]- [5~(0)-\frac{3(0)^2}{2}]+[\frac{3~(3)^2}{2}-5(3)]- [\frac{3(\frac{5}{3})^2}{2}-5(\frac{5}{3})]$ $=(\frac{25}{3}-\frac{25}{6})- (0)+(\frac{27}{2}-15)- (\frac{25}{6}-\frac{25}{3})$ $=(\frac{25}{6})-(\frac{3}{2})+ (\frac{25}{6})$ $=(\frac{25}{6})-(\frac{9}{6})+ (\frac{25}{6})$ $= \frac{41}{6}~m$
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