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: 47

Answer

$$\frac{1}{2}\ln \left| y \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{1}{{2y}}} dy \cr & {\text{take out the constant}} \cr & = \frac{1}{2}\int {\frac{1}{y}} dy \cr & {\text{integrate}} \cr & = \frac{1}{2}\left( {\ln \left| y \right|} \right) + C \cr & = \frac{1}{2}\ln \left| y \right| + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dy}}\left( {\frac{1}{2}\ln \left| y \right| + C} \right) \cr & {\text{ = }}\frac{d}{{dy}}\left( {\frac{1}{2}\ln \left| y \right|} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }}\frac{1}{2}\left( {\frac{1}{y}} \right) + 0 \cr & = \frac{1}{{2y}} \cr} $$
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