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: 32

Answer

$$\frac{52}{3}$$

Work Step by Step

Given $$y=6-x^3,\ \ \ y=x^2-6, \ \ \ y=0 $$ First, we find the intersection points \begin{align*} 6-x^3&=x^2-6\\ x^3+x^2-12&=0\\ (x-2)\left(x^{2}+3 x+6\right)&=0 \end{align*} Then $x= 2$ and $ 6-x^3\geq x^2-6$ for $0\leq x\leq 2$; hence, the area is given by \begin{aligned} \text{Area}&=\int_{0}^{2}\left[\left(6-x^{3}\right)-\left(x^{2}-6\right)\right]d x\\ &=\int_{0}^{2}\left( -x^{3} -x^2+12\right) d x\\ &=\left( \frac{-1}{4}x^{4} -\frac{1}{3}x^3+12x\right) \bigg|_{0}^{2}\\ &=\frac{52}{3} \end{aligned} We use the Mathematica program to plot the region.
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