Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 6 - Review - Exercises - Page 479: 1

Answer

$\frac{64}{3}$

Work Step by Step

Find the intersection between $y=x^2$ and $y=8x-x^2$: $y=y$ $x^2=8x-x^2$ $2x^2-8x=0$ $x(2x-8)=0$ $x=0\vee x=4$ So, both parabola curves intersects at $x=0$ and $x=4$. Describe the region bounded by the curves. $y=x^2$ is a function for an upward parabola and $y=8x-x^2$ is a function for a downward parabola. Graphically, the region is bounded above by the graph $y=8x-x^2$ and bounded below by the graph $y=x^2$ between $x=0$ and $x=4$. Calculate the area of the region: $A=\int_0^4(8x-x^2)-x^2 dx$ $=\int_0^48x-2x^2 dx$ $=[4x^2-\frac{2x^3}{3}]_0^4$ $=(4\cdot 4^2-\frac{2\cdot 4^3}{3})-(4\cdot 0^2-\frac{2\cdot 0^3}{3})$ $=(64-\frac{128}{3})-0$ $=\frac{64}{3}$
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