Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - Review Exercises: 24

Answer

\[ = \frac{{\,{{\left( {3{y^3} + 1} \right)}^5}}}{{45}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {{y^2}\,{{\left( {3{y^3} + 1} \right)}^4}dy} \hfill \\ \hfill \\ set\,\,u = 3{y^3} + 1 \hfill \\ du = 9{y^2}dy \hfill \\ \frac{1}{9}du = {y^2}dy \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \int_{}^{} {{y^2}\,{{\left( {3{y^3} + 1} \right)}^4}dy} = \frac{1}{9}\int {{u^4}du} \hfill \\ \hfill \\ integrate \hfill \\ \frac{1}{9}{u^5} + C \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ = \frac{1}{9}\,\left( {\frac{{\,{{\left( {3{y^3} + 1} \right)}^5}}}{5}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{\,{{\left( {3{y^3} + 1} \right)}^5}}}{{45}} + C \hfill \\ \end{gathered} \]
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