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: 94

Answer

$$\frac{\pi }{6}$$

Work Step by Step

$$\eqalign{ & \int_0^{\sqrt 3 } {\frac{{3dx}}{{9 + {x^2}}}} \cr & {\text{integrate using the rule }}\int {\frac{1}{{{a^2} + {u^2}}}du = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr & = 3\left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^{\sqrt 3 } \cr & = \left[ {{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^{\sqrt 3 } \cr & {\text{using the fundamental theorem}} \cr & = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{3}} \right) - {\tan ^{ - 1}}\left( {\frac{0}{3}} \right) \cr & {\text{simplify}} \cr & = \frac{\pi }{6} - 0 \cr & = \frac{\pi }{6} \cr} $$
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