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: 27

Answer

\[ = \frac{{\,{{\left( {{x^6} - 3{x^2}} \right)}^5}}}{{30}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,{{\left( {{x^6} - 3{x^2}} \right)}^4}\,\left( {{x^5} - x} \right)dx} \hfill \\ \hfill \\ set\,\,the\,\,substitution \hfill \\ \hfill \\ u = {x^6} - 3{x^2}\,\,\,\,\,\,then\,\,\,\,\,\,du = 6{x^5} - 6xdx \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\,{{\left( {{x^6} - 3{x^2}} \right)}^4}\,\left( {{x^5} - x} \right)dx} \, = \frac{1}{6} \cdot \int_{}^{} {\,{{\left( {{x^6} - 3{x^2}} \right)}^4}\,\left( {6{x^5} - 6x} \right)dx} \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \frac{1}{6}\int_{}^{} {{u^4}du} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ \frac{1}{6}.\frac{{{u^5}}}{5} + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,\,u = {x^6} - 3{x^2} \hfill \\ \hfill \\ = \frac{{\,{{\left( {{x^6} - 3{x^2}} \right)}^5}}}{{30}} + C \hfill \\ \end{gathered} \]
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