Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 4

Answer

$ \frac{32}{3} $

Work Step by Step

Step 1 of 3: First, find the points of intersection of the curves $ x=-y $ and $ x=y^{2}+3y $ $ y^{2}+3y=-y $ $ y^{2}+4y=0 $ $ y \left( y+4 \right) =0 $ $ y=0 $ or $ y=-4 $ The y-coordinates of the points of intersection are 0 and -4. Step 2 of 3: Sketch the curves. Find the area of the shaded region. Step 3 of 3: The area $ A= \int _{-4}^{0} \left[ \left( -y \right) - \left( y^{2}+3y \right) \right] dy $ $ A= \int _{-4}^{0} \left( -y-y^{2}-3y \right) dy $ $ A= \int _{-4}^{0} \left( -4y-y^{2} \right) dy $ $ A= \left[ -2y^{2}-\frac{y^{3}}{3} \right] _{-4}^{0} $ $ A= \left[ 0-0- \left( -2 \left( -4 \right) ^{2}-\frac{ \left( -4 \right) ^{3}}{3} \right) \right] $ $ A= \left[ - \left( -32+\frac{64}{3} \right) \right] $ $ A=32-\frac{64}{3} $ $ A=\frac{32}{3} $
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