Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises: 12

Answer

\[ = - \frac{s}{2}{e^{ - 2s}} - \frac{1}{4}{e^{ - 2s}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {s{e^{ - 2s}}ds} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = s\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,du = ds \hfill \\ dv = {e^{ - 2s}}\,\,ds\,\,\,\, \to \,\,\,\,\,v\,\, = - \frac{1}{2}{e^{ - 2s}} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = - \frac{s}{2}{e^{ - 2s}} + \int_{}^{} {\frac{1}{2}{e^{ - 2s}}ds} \hfill \\ then \hfill \\ = - \frac{s}{2}{e^{ - 2s}} - \frac{1}{4}\int_{}^{} {{e^{ - 2s}}\left( { - 2} \right)ds} \hfill \\ \hfill \\ {\text{integrating}} \hfill \\ \hfill \\ = - \frac{s}{2}{e^{ - 2s}} - \frac{1}{4}{e^{ - 2s}} + C \hfill \\ \end{gathered} \]
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