Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.1 Areas Between Curves - 5.1 Exercises - Page 362: 13



Work Step by Step

$y=12-x^2$, $y=x^2-6$ Find their intersection points: $12-x^2=x^2-6$ $2x^2=18$ $x^2=9$ $x=3, -3$ So the area is: $\int_{-3}^3((12-x^2)-(x^2-6))dx$ $=\int_{-3}^3(18-2x^2)dx$ $=(18x-\frac{2x^3}{3})|_{-3}^3$ $=\left(18*3-\frac{2*3^3}{3}\right)-\left(18(-3)-\frac{2(-3)^3}{3}\right)$ $=(54-18)-(-54-(-18))$ $=36-(-36)$ $=72$
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