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: 29


$$\int^{4}_1(\frac{4+6u}{\sqrt u})du=36$$

Work Step by Step

$$A=\int^{4}_1(\frac{4+6u}{\sqrt u})du$$ $$A=\int^{4}_1[({4+6u})u^{-1/2}]du$$ $$A=\int^{4}_1(4u^{-1/2}+6u^{1/2})du$$According to Table 1, we have $$\int(x^n)dx=\frac{x^{n+1}}{n+1}+C (n\ne-1)$$ $$\int cf(x)dx=c\int f(x)$$ Therefore, $$A=(4\frac{u^{1/2}}{\frac{1}{2}}+6\frac{u^{3/2}}{\frac{3}{2}})\Bigg]^4_1$$ $$A=(8u^{1/2}+4u^{3/2})\Bigg]^4_1$$ $$A=(8\sqrt u+4\sqrt{u^3})\Bigg]^4_1$$ $$A=(8\sqrt 4+4\sqrt{4^3})-(8\sqrt 1+4\sqrt{1^3})$$ $$A=(8\times2+4\times8)-(8+4)$$ $$A=36$$
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