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: 28

Answer

\[\frac{1}{3}\,{\left( {{x^2} - 6x} \right)^{3/2}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,\left( {\sqrt {{x^2} - 6x} } \right)\,\left( {x - 3} \right)dx} \hfill \\ Let\,\,u = {x^2} - 6x\,\,,\,\,So\,\,that \hfill \\ du = \,\left( {2x - 6} \right)dx \hfill \\ du = 2\,\left( {x - 3} \right)dx \hfill \\ Then \hfill \\ \frac{1}{2}\int_{}^{} {\,\left( {\sqrt {{x^2} - 6x} } \right)\,\left( {x - 3} \right)dx} = \frac{1}{2}\int_{}^{} {\sqrt u du} \hfill \\ \frac{1}{2}\int_{}^{} {{u^{1/2}}du} \hfill \\ Use\,\,power\,\,rule \hfill \\ \frac{1}{2}\,\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \hfill \\ \frac{1}{3}{u^{3/2}} + C \hfill \\ Substituting\,\,u = {x^2} - 6x\,\,,\,\,gives \hfill \\ \frac{1}{3}\,{\left( {{x^2} - 6x} \right)^{3/2}} + C \hfill \\ \hfill \\ \end{gathered} \]
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.