Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 390: 34

Answer

$${e^{2x - 1}} + C $$

Work Step by Step

$$\eqalign{ & \int {2{e^{2x - 1}}} dx \cr & = 8\int {{e^{x + 1}}} dx \cr & {\text{set }}u = 2x - 1{\text{ then }}\frac{{du}}{{dx}} = 2 \to \frac{{du}}{2} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {2{e^{2x - 1}}} dx = \int {2{e^u}} \left( {\frac{{du}}{2}} \right) \cr & = \int {{e^u}} du \cr & {\text{integrating}} \cr & = {e^u} + C \cr & {\text{replace }}2x - 1{\text{ for }}u \cr & = {e^{2x - 1}} + C \cr} $$
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