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: 10

Answer

$$A=\frac{2\sqrt{t^7}}{7}+\frac{6\sqrt{t^5}}{5}+\frac{4\sqrt{t^3}}{3}+C$$

Work Step by Step

$$A=\int\sqrt t(t^2+3t+2)dt$$ $$A=\int t^{1/2}(t^2+3t+2)dt$$ $$A=\int(t^{5/2}+3t^{3/2}+2t^{1/2})dt$$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, $$A=\int (t^{5/2})dt+3\int(t^{3/2})dt+2\int(t^{1/2})dt$$ From Table 1, we also get that $$\int (x^n)dx=\frac{x^{n+1}}{n+1}$$ Therefore, $$A=\frac{t^{7/2}}{\frac{7}{2}}+\frac{3t^{5/2}}{\frac{5}{2}}+\frac{2t^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{2t^{7/2}}{7}+\frac{6t^{5/2}}{5}+\frac{4t^{3/2}}{3}+C$$ $$A=\frac{2\sqrt{t^7}}{7}+\frac{6\sqrt{t^5}}{5}+\frac{4\sqrt{t^3}}{3}+C$$
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