Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Review - Exercises: 5

Answer

$ \frac{4}{ \pi }+\frac{4}{3} $

Work Step by Step

Step 1 of 3: Find the point of intersection of the curves $ y= \left( sin\frac{ \pi x}{2} \right) $ and $ y= x^{2}-2x $ When$ x= 0 $ , both the curves have value 0. When $ x= 2 $ , both the curves have value 0. The x-coordinates of the points of intersection are 0 and 2. Step 2 of 3: Sketch the curves. Find the area of the shaded region. Step 3 of 3: The area $ A= \int _{0}^{2} \left[ sin \left( \frac{ \pi x}{2} \right) - \left( x^{2}-2x \right) \right] dx $ $ A= \int _{0}^{2}sin \left( \frac{ \pi x}{2} \right) dx- \int _{0}^{2} \left( x^{2}-2x \right) dx $ $ = \left[ -\frac{2}{ \pi }cos \left( \frac{ \pi x}{2} \right) \right] _{0}^{2}- \left[ \frac{x^{3}}{3}-x^{2} \right] _{0}^{2} $ [By the fundamental theorem of calculus, part 2] $ = \left[ -\frac{2}{ \pi }cos \pi +\frac{2}{ \pi }cos \pi \right] - \left[ \frac{8}{3}-4 \right] $ $ = \left[ \frac{2}{ \pi }+\frac{2}{ \pi } \right] - \left[ -\frac{4}{3} \right] =\frac{4}{ \pi }+\frac{4}{3} $ $ A=\frac{4}{ \pi }+\frac{4}{3} $
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