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: 44

Answer

\[ = 2\]

Work Step by Step

\[\begin{gathered} \int_0^4 {\frac{p}{{\sqrt {9 + {p^2}} }}\,\,\,\,dp} \hfill \\ \hfill \\ set\,\,u = 9 + {p^2}\,\,\,\,then\,\,\,du = 2pdp \hfill \\ \hfill \\ {\text{Changing limits of integration}} \hfill \\ \hfill \\ P = 0\,\,\,\,implies\,\,\,\,u = 9 \hfill \\ P = 4\,\,\,\,implies\,\,\,u = 25 \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ = \frac{1}{2}\int_9^{25} {\frac{{du}}{{{u^{\frac{1}{2}}}}}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ {\frac{1}{2}\,\left( {2{u^{\frac{1}{2}}}} \right)} \right]_9^{25} \hfill \\ \hfill \\ = \,\,\left[ {\sqrt u } \right]_9^{25} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \sqrt {25} - \sqrt 9 \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = 2 \hfill \\ \end{gathered} \]
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