Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 6



Work Step by Step

The point of intersection can be found as: $15-2y=y^2-9 \\ 15-2y-y^2+9=0 \\-y^2-2y+24 =0 \\ y=-6 ; 4$ The area $(A)$ can be calculated as: $A=\int_{-6}^{4} [(15-2y)-(y^2-9)] \ dy \\= \int_{-6}^4 [15-2y-y^2+9] \ dy \\=\int_{-6}^4 [-y^2-2y+24] \ dy \\=[-\dfrac{y^3}{3}-y^2+24y]_{-6}^{4} \\= 48-24 \\=\dfrac{500}{3}$
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