Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 86

Answer

$$\frac{1}{2}$$

Work Step by Step

$$\eqalign{ & \frac{1}{2}\int_0^{\ln 2} {{e^x}} dx \cr & \frac{d}{{dx}}\left[ {{e^x}} \right] = {e^x},{\text{ then}} \cr & \frac{1}{2}\int_0^{\ln 2} {{e^x}} dx = \frac{1}{2}\left. {\left( {{e^x}} \right)} \right|_0^{\ln 2} \cr & {\text{using the fundamental theorem}} \cr & = \frac{1}{2}\left( {{e^{\ln 2}} - {e^0}} \right) \cr & {\text{simplify}} \cr & = \frac{1}{2}\left( {2 - 1} \right) \cr & = \frac{1}{2} \cr} $$
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