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: 34

Answer

\[ = 2\ln 2 - 1\]

Work Step by Step

\[\begin{gathered} \int_0^{\ln 2} {x{e^x}dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = dx \hfill \\ dv = {e^x}dx\,\,\,\,\, \to \,\,\,\,\,\,\,v = {e^x} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = x{e^x} - \int_{}^{} {{e^x}dx} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ x{e^x} - {e^x} + C \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ \int_0^{\ln 2} {x{e^x}dx} = \,\,\left[ {x{e^x} - {e^x}} \right]_0^{\ln 2} \hfill \\ \hfill \\ evaluate\,\,the\,\,limits \hfill \\ \hfill \\ = \,\left( {\ln 2{e^{\ln 2}} - {e^{\ln 2}}} \right) - \,\left( {0 - {e^0}} \right) \hfill \\ \hfill \\ = 2\ln 2 - 2 + 1 \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 2\ln 2 - 1 \hfill \\ \hfill \\ \end{gathered} \]
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