University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 96


$$\int^8_1\Big(\frac{2}{3x}-\frac{8}{x^2}\Big)dx=\ln 4-7$$

Work Step by Step

$$A=\int^8_1\Big(\frac{2}{3x}-\frac{8}{x^2}\Big)dx$$ $$A=\frac{2}{3}\int^8_1\frac{1}{x}dx+8\int^8_1-\frac{1}{x^2}dx$$ $$A=\frac{2}{3}(\ln x)\Big]^8_1+8\times\Big(\frac{1}{x}\Big)\Big]^8_1$$ $$A=\frac{2}{3}(\ln8-\ln1)+8\Big(\frac{1}{8}-1\Big)$$ $$A=\frac{2\ln8}{3}+1-8$$ $$A=\frac{2\ln8}{3}-7$$ $$A=\frac{2\ln2^3}{3}-7$$ $$A=2\ln2-7$$ $$A=\ln4-7$$
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