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 - Page 374: 54

Answer

$$A = 18$$

Work Step by Step

$$\eqalign{ & \left( i \right){\text{From the graph we can see that the net area of the region is:}} \cr & A = \int_{ - \pi /2}^{\pi /2} {6\cos x} dx - \int_{\pi /2}^\pi {6\cos x} dx \cr & {\text{Integrating}} \cr & A = 6\left[ {\sin x} \right]_{ - \pi /2}^{\pi /2} - 6\left[ {\sin x} \right]_{\pi /2}^\pi \cr & {\text{Evaluating}} \cr & A = 6\left[ {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( { - \frac{\pi }{2}} \right)} \right] - 6\left[ {\sin \left( \pi \right) - \sin \left( {\frac{\pi }{2}} \right)} \right] \cr & A = 6\left( 2 \right) - 6\left[ {\sin \left( \pi \right) - \sin \left( {\frac{\pi }{2}} \right)} \right] \cr & A = 12 + 6 \cr & A = 18 \cr} $$
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