Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 40

Answer

\[\frac{\pi }{6}\]

Work Step by Step

\[\begin{gathered} \int_0^{1/2} {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \hfill \\ \hfill \\ {\text{integrating}}{\text{, }}\frac{d}{{dx}}\left[ {\arcsin x} \right] = \frac{1}{{\sqrt {1 - {x^2}} }} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_0^{1/2} {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \left[ {{{\sin }^{ - 1}}\left( x \right)} \right]_0^{1/2} \hfill \\ \hfill \\ {\text{Use the fundamental theorem of calculus}} \hfill \\ \hfill \\ {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) - {\sin ^{ - 1}}\left( 0 \right) \hfill \\ \hfill \\ \frac{\pi }{6} \hfill \\ \end{gathered} \]
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