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: 32

Answer

$$\int^4_1\frac{\sqrt y-y}{y^2}dy=1-\ln4$$

Work Step by Step

$$A=\int^4_1\frac{\sqrt y-y}{y^2}dy$$ $$A=\int^4_1\frac{y^{1/2}-y}{y^2}dy$$ $$A=\int^4_1(y^{-3/2}-\frac{1}{y})dy$$ 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, $$A=(\frac{y^{-1/2}}{-\frac{1}{2}}-\ln|y|)\Bigg]^4_1$$ $$A=(-\frac{2}{\sqrt y}-\ln|y|)\Bigg]^4_1$$ $$A=(-\frac{2}{\sqrt 4}-\ln|4|)-(-\frac{2}{\sqrt 1}-\ln|1|)$$ $$A=(-1-\ln4)+(2+0)$$ $$A=1-\ln4$$
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