Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - Chapter Review - Review Exercises - Page 417: 27

Answer

$$\frac{2}{{{x^2}}} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{ - 4}}{{{x^3}}}} dx \cr & {\text{use the property of exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr & = \int {\left( { - 4{x^{ - 3}}} \right)} dx \cr & {\text{use multiple constant rule }}\int {k \cdot f\left( x \right)} dx = k\int {f\left( x \right)} dx \cr & = - 4\int {{x^{ - 3}}} dx \cr & {\text{use power rule }}\int {{x^x}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & = - 4\left( {\frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right) + C \cr & {\text{simplifying}} \cr & = - 4\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right) + C \cr & = 2{x^{ - 2}} + C \cr & = \frac{2}{{{x^2}}} + C \cr} $$
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