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: 8

Answer

$$A=\frac{u^7}{7}-\frac{u^6}{3}-\frac{u^4}{4}+\frac{2u}{7}+C$$

Work Step by Step

$$A=\int(u^6-2u^5-u^3+\frac{2}{7})du$$ From Table 1, $$\int cf(x)dx=c\int f(x)dx$$ $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ Therefore, $$A=\int(u^6)du-2\int(u^5)du-\int(u^3)du+\int\frac{2}{7}du$$ From Table 1, we also get the followings $$\int kdx=kx+C$$ $$\int (x^n)dx=\frac{x^{n+1}}{n+1}$$ Therefore, $$A=\frac{u^7}{7}-2\frac{u^6}{6}-\frac{u^4}{4}+\frac{2u}{7}+C$$ $$A=\frac{u^7}{7}-\frac{u^6}{3}-\frac{u^4}{4}+\frac{2u}{7}+C$$
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