Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 329: 109

Answer

$$\frac{4}{{15}}{x^{15/2}} - \frac{{24}}{{11}}{x^{11/6}} + C$$

Work Step by Step

$$\eqalign{ & \int {\sqrt x \left( {2{x^6} - 4\root 3 \of x } \right)dx} \cr & {\text{rewrite the radicals}} \cr & = \int {{x^{1/2}}\left( {2{x^6} - 4{x^{1/3}}} \right)dx} \cr & {\text{multiply}} \cr & = \int {\left( {2{x^{13/2}} - 4{x^{5/6}}} \right)dx} \cr & {\text{integrate by the power rule}} \cr & = 2\left( {\frac{{{x^{15/2}}}}{{15/2}}} \right) - 4\left( {\frac{{{x^{11/6}}}}{{11/6}}} \right) + C \cr & = \frac{4}{{15}}{x^{15/2}} - \frac{{24}}{{11}}{x^{11/6}} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {\frac{4}{{15}}{x^{15/2}} - \frac{{24}}{{11}}{x^{11/6}} + C} \right) \cr & {\text{ = }}\frac{d}{{dx}}\left( {\frac{4}{{15}}{x^{15/2}}} \right) - \frac{d}{{dx}}\left( {\frac{{24}}{{11}}{x^{11/6}}} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }}\frac{4}{{15}}\left( {\frac{{15}}{2}} \right){x^{13/2}} - \frac{{24}}{{11}}\left( {\frac{{11}}{6}} \right){x^{5/6}} + 0 \cr & {\text{simplify}} \cr & {\text{ = }}2{x^{13/2}} - 4{x^{5/6}} \cr} $$
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