Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - Review Exercises: 20

Answer

\[ = \frac{{{e^{16}} - 1}}{4}\]

Work Step by Step

\[\begin{gathered} \int_{ - 2}^2 {{e^{4x + 8}}dx} \hfill \\ \hfill \\ set\,\,u = 4x + 8{\text{ then}}\,{\text{ }}du = 4dx \hfill \\ \frac{1}{4}du = dx \hfill \\ \hfill \\ = \int_{}^{} {{e^{4x + 8}}\,} \,dx = \frac{1}{4}\int {{e^u}} du \hfill \\ integrate \hfill \\ = \frac{1}{4}{e^u} + C \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{ - 2}^2 {{e^{4x + 8}}dx} = \frac{1}{4}\,\,\left[ {{e^{4x + 8}}} \right]_{ - 2}^2 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{1}{4}\,\,\left[ {{e^{4\,\left( 2 \right) + 8}} - {e^{4\,\left( { - 2} \right) + 8}}} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{1}{4}\,\left( {{e^{16}} - {e^0}} \right) \hfill \\ \hfill \\ = \frac{{{e^{16}} - 1}}{4} \hfill \\ \end{gathered} \]
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