Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 288: 66

Answer

a. $1\ ft$, $10.17\ ft$, and $13\ ft$, b. $\frac{29}{3}\approx9.67\ ft$

Work Step by Step

a. Given $H(t)=\sqrt {t+1}+5t^{1/3}=(t+1)^{1/2}+5t^{1/3}$ for $0\leq t\leq8$, we have $H(0)=\sqrt {0+1}+5(0)^{1/3}=1\ ft$, $H(4)=\sqrt {4+1}+5(4)^{1/3}=\sqrt 5+5\sqrt[3] 4\approx10.17\ ft$, and $H(8)=\sqrt {8+1}+5(8)^{1/3}=3+10=13\ ft$, b. The average height can be calculated as $\bar H=\frac{1}{8-0}\int_0^8H(t)dt=\frac{1}{8}\int_0^8((t+1)^{1/2}+5t^{1/3})dt=\frac{1}{8}(\frac{2}{3}(t+1)^{3/2}+\frac{15}{4}t^{4/3})|_0^8=\frac{1}{8}(\frac{2}{3}(8+1)^{3/2}+\frac{15}{4}(8)^{4/3})-\frac{1}{8}(\frac{2}{3}(0+1)^{3/2}+\frac{15}{4}(0)^{4/3})=\frac{39}{4}-\frac{1}{12}=\frac{29}{3}\approx9.67\ ft$
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