Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 33



Work Step by Step

$$A=\int^2_1(\frac{x}{2}-\frac{2}{x})dx$$ According to Table 1, we have $$\int x^ndx=\frac{x^{n+1}}{n+1}+C (n\ne-1)$$ $$\int \frac{1}{x}dx=\ln|x|+C$$ Therefore, $$A=(\frac{1}{2}\frac{x^2}{2}-2\ln|x|)\Bigg]^2_1$$ $$A=(\frac{x^2}{4}-2\ln|x|)\Bigg]^2_1$$ $$A=(\frac{2^2}{4}-2\ln|2|)-(\frac{1^2}{4}-2\ln|1|)$$ $$A=(1-2\ln2)-(\frac{1}{4}-2\times0)$$ $$A=1-2\ln2-\frac{1}{4}$$ $$A=\frac{3}{4}-2\ln2$$
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