Calculus (3rd Edition)

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

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 10


$$ (\bar{x},\bar{y})=(1,2) $$

Work Step by Step

Since \begin{aligned} M_{y} &=\delta \int_{a}^{b} x f(x) d x \\ &=\int_{0}^{3} x(6-2 x) d x \\ &=\int_{0}^{3}\left(6 x-2 x^{2}\right) d x \\ &=\left.\left(3 x^{2}-\frac{2}{3} x^{3}\right)\right|_{0} ^{3} \\ &=9 \end{aligned} and \begin{aligned} M_{x} &=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{3}(6-2 x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{3}\left(36-24 x+4 x^{2}\right) d x \\ &=\left.\frac{1}{2}\left(36 x-12 x^{2}+\frac{4}{3} x^{3}\right)\right|_{0} ^{3} \\ &=18 \end{aligned} and $$M=\delta A=\int_{0}^{3}(6-2 x) d x=\left.\left(6 x-x^{2}\right)\right|_{0} ^{3}=9$$ Then $$ \bar{x}=\frac{M_{y}}{M}=\frac{9}{9}=1 \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{18}{9}=2 $$
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