Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - Review Exercises: 25

Answer

\[ = 1\]

Work Step by Step

\[\begin{gathered} \int_0^3 {\frac{x}{{\sqrt {25 - {x^2}} }}\,\,dx} \hfill \\ \hfill \\ rewrite\,\, \hfill \\ \hfill \\ = \frac{1}{2}\int_0^3 {\,{{\left( {25 - {x^2}} \right)}^{ - \frac{1}{2}}}\,\left( { - 2x} \right)dx} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = - \frac{1}{2}\,\,\left[ {\frac{{\,\left( {25 - {x^2}} \right)}}{{\frac{1}{2}}}} \right]_0^3 \hfill \\ \hfill \\ = - \,\,\left[ {\sqrt {25 - {x^2}} } \right]_0^3 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = - \sqrt {25 - \,{{\left( 3 \right)}^2}} + \sqrt {25 - {0^2}} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = - 4 + 5 \hfill \\ \hfill \\ = 1 \hfill \\ \hfill \\ \end{gathered} \]
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