Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 17


$2.7552 \ kg$

Work Step by Step

The total mass $(M)$ of the rod can be calculated as: $Mass, M= \int_{0}^{1.2} (1+2x+\dfrac{2x^3}{9}) \ dx \\=[x+x^2+\dfrac{x^4}{18}]_0^{1.2} \\=1.2+(1.2)^2+\dfrac{(1.2)^4}{18} \approx 2.7552 \ kg$
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