Thomas' Calculus 13th Edition

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

Chapter 6: Applications of Definite Integrals - Practice Exercises - Page 363: 26


$22,800,00 $ ft- lb

Work Step by Step

The force can be found as: $F(x)= 8 \times 800 [\dfrac{(2) \times 4750-x}{(2) \times (4750)}] =6400 (1- \dfrac{x}{9500} ) lb$ Integrate the integral to calculate the work done area as follows: Work done, $W=\int_c^d F(x) dx= \int_{0}^{4750}6400 (1-\dfrac{x}{9500}) dx$ or, $=6400[4750-\dfrac{(4750)^2}{4 \times 4750 }]$ or, $=22,800,00 $ ft- lb
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