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 288: 53


$c=\dfrac{\sqrt[3] 9}{4}$

Work Step by Step

The area of under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by: $Area, A= \int_{-\sqrt c}^{\sqrt c} (c-x^2-x^2+c) \ dx \\ = [2cx -\dfrac{2x^3}{3}]_{-\sqrt c}^{\sqrt c} \\=(2c^{3/2} -\dfrac{2 c^{3/2}}{3})-(-2c^{3/2} +\dfrac{2 c^{3/2}}{3}) \\=\dfrac{8 c^{3/2}}{3}$ Since, $Area= 1$ $\dfrac{8 c^{3/2}}{3} =1 \implies c=\dfrac{\sqrt[3] 9}{4}$
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