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: 102

Answer

$$\root 3 \of {{x^2}} + \sqrt {{x^3}} + C$$

Work Step by Step

$$\eqalign{ & \int {\left( {\root 3 \of {{x^2}} + \sqrt {{x^3}} } \right)dx} \cr & {\text{rewritting radicals}} \cr & = \int {\left( {{x^{2/3}} + {x^{3/2}}} \right)dx} \cr & {\text{sum rule}} \cr & = \int {{x^{2/3}}dx} + \int {{x^{3/2}}dx} \cr & {\text{by the power rule}} \cr & = \frac{{{x^{5/3}}}}{{5/3}} + \frac{{{x^{5/2}}}}{{5/2}} + C \cr & simplify \cr & = \frac{3}{5}{x^{5/3}} + \frac{2}{5}{x^{5/2}} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {\frac{3}{5}{x^{5/3}} + \frac{2}{5}{x^{5/2}} + C} \right) \cr & {\text{ = }}\frac{3}{5}\left( {\frac{5}{3}} \right){x^{2/3}} + \frac{2}{5}\left( {\frac{5}{2}} \right){x^{3/2}} + C \cr & {\text{simplify}} \cr & {\text{ = }}{x^{2/3}} + {x^{3/2}} + C \cr & = \root 3 \of {{x^2}} + \sqrt {{x^3}} + C \cr} $$
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