Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 287: 41


$\dfrac{32}{3}$ $units^2$

Work Step by Step

The area under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by: $Area, A= \int_{0}^{4} [2y-((y-1)^2-1)] \ dy \\ = \int_{0}^{4} (2y-y^2+2y) \ dy\\= [2y^2 -\dfrac{1}{3} y^3]_0^4 \\= 2(4^2) -\dfrac{1}{3} (4^3) \\=\dfrac{32}{3}$
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