Calculus with Applications (10th Edition)

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

Chapter 7 - Integration - 7.2 Substitution - 7.2 Exercises: 16

Answer

\[{e^{\sqrt y }} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{e^{\sqrt y }}}}{{2\sqrt y }}dy} \hfill \\ Let\,\,u = \sqrt y = {y^{\frac{1}{2}}} \hfill \\ So\,\,that \hfill \\ du = \frac{1}{2}{y^{ - 1/2}} = \frac{1}{{2\sqrt y }} \hfill \\ Then\,\,\int_{}^{} {\frac{{{e^{\sqrt y }}}}{{2\sqrt y }} = \int_{}^{} {{e^u}du} } \hfill \\ Integrating\, \hfill \\ {e^u} + C \hfill \\ Substituting\,\,u = \sqrt y \,\,for\,\,u\,\,gives\,\, \hfill \\ {e^{\sqrt y }} + C \hfill \\ \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.