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: 15


$$(\bar{x},\bar{y})=\left(\frac{1-5 e^{-4}}{1-e^{-4}},\frac{1-e^{-8}}{4-4 e^{-4}}\right)$$

Work Step by Step

Since \begin{align*} M_{y}&=\delta \int_{a}^{b} x f(x) d x\\ &=\int_{0}^{4} x e^{-x} d x\\ &=-xe^{-x}\bigg|_{0}^{4}+ \int_{0}^{4} e^{-x} d x\\ &=-e^{-x}(x+1)\bigg|_{0}^{4}\\ &= 1-5e^{-4} \end{align*} and \begin{aligned} M_{x}=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x &=\frac{1}{2} \int_{0}^{4} e^{-2 x} d x \\ &=\left.\frac{1}{2}\left(-\frac{1}{2} e^{-2 x}\right)\right|_{0} ^{4} \\ &=\frac{1}{2}\left(-\frac{1}{2} e^{-8}+\frac{1}{2}\right) \\ &=\frac{1}{4}\left(1-e^{-8}\right) \end{aligned} and \begin{align*} M&=\delta A\\ &=\int_{0}^{4} e^{-x} d x\\ &=\left.\left(-e^{-x}\right)\right|_{0} ^{4}=\left(1-e^{-4}\right) \end{align*} Then $$\bar{x}=\frac{M_{y}}{M}=\frac{1-5 e^{-4}}{1-e^{-4}} \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{1-e^{-8}}{4-4 e^{-4}}$$
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