Calculus with Applications (10th Edition)

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

Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises: 12

Answer

\[4{y^4} + 3{y^3} - 3{y^2} + 3y + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,\left( {16{y^3} + 9{y^2} - 6y + 3} \right)dy} \hfill \\ Extending\,\,using\,\,the\,\,sum\,\,and \hfill \\ difference\,\,rules \hfill \\ \int_{}^{} {13{y^3}dy} + \int_{}^{} {9{y^2}dy} - \int_{}^{} {6ydy} + \int_{}^{} {3dy} \hfill \\ Use\,\,the\,\,power\,\,rule \hfill \\ \int_{}^{} {{y^n}dy} = \frac{{{y^{n + 1}}}}{{n + 1}} + C \hfill \\ 16\,\left( {\frac{{{y^4}}}{4}} \right) + 9\,\left( {\frac{{{y^3}}}{3}} \right) - 6\,\left( {\frac{{{y^2}}}{2}} \right) + 3y + C \hfill \\ Simplifying \hfill \\ 4{y^4} + 3{y^3} - 3{y^2} + 3y + C \hfill \\ \end{gathered} \]
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