Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - Exercises - Page 423: 58

Answer

(a) $\frac{175}{6}$ (b) $\frac{59}{2}$

Work Step by Step

(a) We can find the displacement: $\int_{0}^{5}(t^2-t)~dt$ $= (\frac{t^3}{3}-\frac{t^2}{2})~\vert_{0}^{5}$ $= [\frac{(5)^3}{3}-\frac{(5)^2}{2}]-[\frac{(0)^3}{3}-\frac{(0)^2}{2}]$ $= (\frac{125}{3}-\frac{25}{2})-(0)$ $= \frac{250}{6}-\frac{75}{6}$ $= \frac{175}{6}$ (b) $\int_{0}^{5}\vert~v(t)~\vert~dt~~$ is the distance traveled by the particle. On the interval $0 \lt t \lt 1$, the function $t^2-t$ is negative. On the interval $1 \lt t \lt 5$, the function $t^2-t$ is positive. We can find the distance traveled by the particle: $\int_{0}^{5}\vert t^2-t\vert~dt$ $= \int_{0}^{1} (t-t^2)~dt+\int_{1}^{5} (t^2-t)~dt$ $= (\frac{t^2}{2}-\frac{t^3}{3})~\vert_{0}^{1}+(\frac{t^3}{3}-\frac{t^2}{2})~\vert_{1}^{5}$ $= [\frac{(1)^2}{2}-\frac{(1)^3}{3}]-[\frac{(0)^2}{2}-\frac{(0)^3}{3}]+[\frac{(5)^3}{3}-\frac{(5)^2}{2}]-[\frac{(1)^3}{3}-\frac{(1)^2}{2}]$ $= (\frac{1}{2}-\frac{1}{3})-(0)+ (\frac{125}{3}-\frac{25}{2})- (\frac{1}{3}-\frac{1}{2})$ $= \frac{1}{6}+\frac{250}{6}-\frac{75}{6}+\frac{1}{6}$ $= \frac{177}{6}$ $= \frac{59}{2}$
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