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: 12


$$ \left(\bar{x},\bar{y}\right)= \left(\frac{4}{5},\frac{2}{7}\right) $$

Work Step by Step

Since \begin{aligned} M_{y} &=\delta \int_{a}^{b} x f(x) d x \\ &=\int_{0}^{1} x\left(x^{3}\right) d x \\ &=\int_{0}^{1} x^{4} d x \\ &=\left.\left(\frac{x^{5}}{5}\right)\right|_{0} ^{1} \\ &=\frac{1}{5} \end{aligned} and \begin{aligned} M_{x} &=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{1}\left(x^{3}\right)^{2} d x \\ &=\frac{1}{2} \int_{0}^{1} x^{6} d x \\ &=\left.\frac{1}{2}\left(\frac{x^{7}}{7}\right)\right|_{0} ^{1} \\ &=\frac{1}{14} \end{aligned} and $$M=\delta A=\int_{0}^{1} x^{3} d x=\left.\left(\frac{x^{4}}{4}\right)\right|_{0} ^{1}=\frac{1}{4}$$ Then $$\bar{x}=\frac{M_{y}}{M}=\frac{4}{5} \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{2}{7}$$
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