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: 56

Answer

$$2{x^{11}} - 2{e^{12x}} + C$$

Work Step by Step

$$\eqalign{ & \int {\left( {22{x^{10}} - 24{e^{12x}}} \right)dx} \cr & {\text{sum rule}} \cr & = \int {22{x^{10}}dx} - \int {24{e^{12x}}dx} \cr & = 22\int {{x^{10}}dx} - 24\int {{e^{12x}}dx} \cr & {\text{integrate}} \cr & = 22\left( {\frac{{{x^{11}}}}{{11}}} \right) - 24\left( {\frac{1}{{12}}{e^{12x}}} \right) + C \cr & {\text{simplify}} \cr & = 2{x^{11}} - 2{e^{12x}} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {2{x^{11}} - 2{e^{12x}} + C} \right) \cr & {\text{ = }}\frac{d}{{dx}}\left( {2{x^{11}}} \right) - \frac{d}{{dx}}\left( {2{e^{12x}}} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }}2\left( {11} \right){x^{10}} - 2\left( {12} \right){e^{12x}} + 0 \cr & = 22{x^{10}} - 24{e^{12x}} \cr} $$
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