Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.2 Substitution - 7.2 Exercises: 13

Answer

\[\frac{1}{2}{e^{2t - {t^2}}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,\left( {1 - t} \right){e^{2t - {t^2}}}dt} \hfill \\ Let\,\,u = 2t - {t^2}\,\,\,so\,\,that \hfill \\ du = 2 - 2t \hfill \\ du = 2\,\left( {1 - t} \right)dt \hfill \\ Then \hfill \\ \frac{1}{2}\int_{}^{} 2 \,\left( {1 - t} \right){e^{2t - {t^2}}}dt \hfill \\ \frac{1}{2}\int_{}^{} {{e^u}du} \hfill \\ Integrating \hfill \\ \frac{1}{2}{e^u} + C \hfill \\ Substituting\,\,u = 2t - {t^2}\,\,\,gives \hfill \\ \frac{1}{2}{e^{2t - {t^2}}} + C \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.