Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 68

Answer

\[ = - \frac{1}{3}{p^2}{e^{ - 3p}} - \frac{2}{9}p{e^{ - 3p}} - \frac{4}{{27}}{e^{ - 3p}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {{p^2}{e^{ - 3p}}dp} \hfill \\ \hfill \\ Integrate\,\,using\,\,the\,\,formula\, \hfill \\ \hfill \\ \int_{}^{} {{x^n}{e^{ax}}dx} = \frac{1}{a}{x^n}{e^{ax}} - \frac{n}{a}\int_{}^{} {{x^{n - 1}}{e^{ax}}dx} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {{p^2}{e^{ - 3p}}dp} = - \frac{1}{3}{p^2}{e^{ - 3p}} - \frac{2}{{ - 3}}\int_{}^{} {x{e^{ - 3p}}dp} \hfill \\ \hfill \\ = - \frac{1}{3}{p^2}{e^{ - 3p}} + \frac{2}{3}\,\,\left[ {\frac{1}{{ - 3}}p{e^{ - 3p}} - \frac{1}{3}\int_{}^{} {{e^{ - 3p}}dp} } \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = - \frac{1}{3}{p^2}{e^{ - 3p}} - \frac{2}{9}p{e^{ - 3p}} - \frac{4}{{27}}{e^{ - 3p}} + C \hfill \\ \end{gathered} \]
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