Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 14

Answer

$G\left( x \right) = \frac{4}{3}{x^{3/2}} - \frac{{2{x^{5/2}}}}{5} + \frac{{12{x^{7/2}}}}{7} + C$

Work Step by Step

$$\eqalign{ & g\left( x \right) = \sqrt x \left( {2 - x + 6{x^2}} \right) \cr & {\text{First use the radical property }}\sqrt m = {m^{1/2}} \cr & g\left( x \right) = {x^{1/2}}\left( {2 - x + 6{x^2}} \right) \cr & {\text{Use the distributive property}} \cr & g\left( x \right) = 2{x^{1/2}} - {x^{3/2}} + 6{x^{5/2}} \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr & {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr & G\left( x \right) = 2\left( {\frac{{{x^{1/2 + 1}}}}{{1/2 + 1}}} \right) - \frac{{{x^{3/2 + 1}}}}{{3/2 + 1}} + 6\left( {\frac{{{x^{5/2 + 1}}}}{{5/2 + 1}}} \right) + C \cr & G\left( x \right) = 2\left( {\frac{{{x^{3/2}}}}{{3/2}}} \right) - \frac{{{x^{5/2}}}}{{5/2}} + 6\left( {\frac{{{x^{7/2}}}}{{7/2}}} \right) + C \cr & G\left( x \right) = \frac{4}{3}{x^{3/2}} - \frac{{2{x^{5/2}}}}{5} + \frac{{12{x^{7/2}}}}{7} + C \cr} $$
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