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 - Page 392: 80

Answer

\[ = \frac{1}{2}\]

Work Step by Step

\[\begin{gathered} f\,\left( \theta \right) = \cos \theta \sin \theta \hfill \\ f\,\left( \theta \right) = 0\,\,\,,\,\,\,\,\left( {axis - \theta } \right) \hfill \\ \hfill \\ Let\,\,\theta = 0\,\,\,\,\,\,and\,\,\,\,\theta = \frac{\pi }{2} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ A = \int_0^{\frac{\pi }{2}} {\cos \theta \sin \theta d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ {\frac{{{{\sin }^2}\theta }}{2}} \right]_0^{\frac{\pi }{2}} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{{{{\sin }^2}\,\left( {\frac{\pi }{2}} \right)}}{2} - \frac{{{{\sin }^2}\,\left( 0 \right)}}{2} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{\,{{\left( 1 \right)}^2}}}{2} - \frac{{{0^2}}}{2} \hfill \\ \hfill \\ = \frac{1}{2} \hfill \\ \hfill \\ \end{gathered} \]
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