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: 42

Answer

$$\int^{2}_{1}\frac{(x-1)^3}{x^2}dx=-2+3\ln2$$

Work Step by Step

$$A=\int^{2}_{1}\frac{(x-1)^3}{x^2}dx$$ $$A=\int^{2}_{1}\frac{x^3-3x^2+3x-1}{x^2}dx$$ $$A=\int^{2}_{1}(x-3+\frac{3}{x}-x^{-2})dx$$ According to Table 1, we have $$\int x^ndx=\frac{x^{n+1}}{n+1}+C(n\ne-1)$$ $$\int kdx=kx+C$$ $$\int\frac{1}{x}dx=\ln|x|+C$$ Therefore, $$A=(\frac{x^2}{2}-3x+3\ln|x|-\frac{x^{-1}}{-1})\Bigg]^{2}_{1}$$ $$A=(\frac{x^2}{2}-3x+3\ln|x|+\frac{1}{x})\Bigg]^{2}_{1}$$ $$A=(\frac{2^2}{2}-3\times2+3\ln|2|+\frac{1}{2})-(\frac{1^2}{2}-3\times1+3\ln|1|+\frac{1}{1})$$ $$A=(-\frac{7}{2}+3\ln2)-(-\frac{3}{2}+3\times0)$$ $$A=-\frac{7}{2}+3\ln2+\frac{3}{2}$$ $$A=-2+3\ln2$$
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