Calculus with Applications (10th Edition)

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

Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises: 25

Answer

\[6{t^{ - 1.5}} - 2\ln \left| t \right| + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,\left( { - 9{t^{ - 2.5}} - 2{t^{ - 1}}} \right)dt} \hfill \\ Use\,\,the\,\,sum\,\,and\,\,difference\,\,rules \hfill \\ \int_{}^{} { - 9{t^{ - 2.5}}dt} - \int_{}^{} {\,\left( { - 2{t^{ - 1}}} \right)} dt \hfill \\ - 9\int_{}^{} {{t^{ - 2.5}}dt} - 2\int_{}^{} {\frac{1}{t}dt} \hfill \\ Integrate\,\,use\,\,the\,\,power\,\,rule\,\,\,and \hfill \\ \int_{}^{} {\frac{1}{x}dx} = \ln \left| x \right| + C \hfill \\ Then \hfill \\ - 9\,\left( {\frac{{{t^{ - 2.5 + 1}}}}{{ - 2.5 + 1}}} \right) - 2\,\left( {\ln \left| t \right|} \right) + C \hfill \\ Simplifying \hfill \\ - 9\,\left( {\frac{{{t^{ - 1.5}}}}{{ - 1.5}}} \right) - 2\ln \left| t \right| + C \hfill \\ 6{t^{ - 1.5}} - 2\ln \left| t \right| + C \hfill \\ \end{gathered} \]
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