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: 34

Answer

\[\frac{2}{3}{y^{\frac{1}{2}}} - \frac{1}{4}{y^2} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{2{y^{\frac{1}{2}}} - 3{y^2}}}{{6y}}dy} \hfill \\ rewrite\,\,the\,\,integrand\,\,as\,\,follows \hfill \\ \int_{}^{} \begin{gathered} \frac{{2{y^{\frac{1}{2}}}}}{{6y}}dy - \int_{}^{} {\frac{{3{y^2}}}{{6y}}dy} \hfill \\ Simplifying \hfill \\ \end{gathered} \hfill \\ \frac{1}{3}\int_{}^{} {{y^{ - \frac{1}{2}}}dy - \frac{1}{2}\int_{}^{} {ydy} } \hfill \\ Use\,\,power\,\,rule\,\,\int_{}^{} {{y^n}dy} = \frac{{{y^{n + 1}}}}{{n + 1}} + C \hfill \\ \frac{1}{3}\,\left( {\frac{{{1^{\frac{1}{2}}}}}{{1/2}}} \right) - \frac{1}{2}\,\left( {\frac{{{y^2}}}{2}} \right) + C \hfill \\ Then \hfill \\ \frac{2}{3}{y^{\frac{1}{2}}} - \frac{1}{4}{y^2} + C \hfill \\ \end{gathered} \]
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