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: 24

Answer

\[\frac{8}{5}\,{\left( {8 - r} \right)^{5/2}} - \frac{{64}}{3}\,{\left( {8 - r} \right)^{3/2}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {4r\sqrt {8 - r} dr} \hfill \\ Let\,\,u = 8 - r\,\,Then\,\,r = 8 - u \hfill \\ So\,\,that\,\,dr = - du \hfill \\ \int_{}^{} {4r\sqrt {8 - r} \,dr = 4\int_{}^{} {\,\left( {8 - u} \right){u^{\frac{1}{2}}}\,\left( { - 1} \right)du} } \hfill \\ = 4\int_{}^{} {\,\left( {u - 8} \right){u^{\frac{1}{2}}}du} \hfill \\ = 4\int_{}^{} {\,\left( {{u^{3/2}} - 8{u^{\frac{1}{2}}}} \right)du} \hfill \\ Power\,\,rule \hfill \\ 4\,\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) - 32\,\left( {\frac{{{u^{\frac{3}{2}}}}}{{3/2}}} \right) + C \hfill \\ Substituting\,\,u = 8 - r\,\,gives \hfill \\ \frac{8}{5}\,{\left( {8 - r} \right)^{5/2}} - \frac{{64}}{3}\,{\left( {8 - r} \right)^{3/2}} + C \hfill \\ \end{gathered} \]
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