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: 48

Answer

\[ = \frac{{4\sqrt {10} - 4}}{3}\]

Work Step by Step

\[\begin{gathered} \int_0^3 {\frac{{{v^2} + 1}}{{\sqrt {{v^3} + 3v + 4} }}} \,\,dv \hfill \\ \hfill \\ set\,\,u = {v^3} + 3v + 4\,\,\,\,then\,\,\,du = 3\,\left( {{v^2} + 1} \right)dv \hfill \\ \hfill \\ v = 0\,\,\,\,implies\,\,\,u = 4 \hfill \\ v = 3\,\,\,\,\,implies\,\,\,u = 40 \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ = \frac{1}{3}\int_4^{40} {\frac{{du}}{{\sqrt u }}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{2}{3}\,\,\,\left[ {\sqrt u } \right]_4^{40} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{2}{3}\,\left( {\sqrt {40} - \sqrt 4 } \right) \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \frac{{4\sqrt {10} - 4}}{3} \hfill \\ \end{gathered} \]
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