Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 43

Answer

\[ = \frac{{{e^9} - 1}}{3}\]

Work Step by Step

\[\begin{gathered} \int_{ - 1}^2 {{x^2}{e^{{x^3} + 1}}dx} \hfill \\ \hfill \\ set\,\,u = {x^3} + 1\,\,\,{\text{which implies that}}\,\,\,du = 3{x^2}dx \hfill \\ \hfill \\ {\text{Changing limits of integration}} \hfill \\ \hfill \\ x = - 1\,\,\,implies\,\,u = 0 \hfill \\ x = 2\,\,\,\,\,implies\,\,u = 9 \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ = \int_0^9 {{e^u}du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \frac{1}{3}\,\,\left[ {{e^u}} \right]_0^9 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{1}{3}\,\left( {{e^9} - {e^0}} \right) \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \frac{{{e^9} - 1}}{3} \hfill \\ \end{gathered} \]
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