Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.1 Areas Between Curves - 5.1 Exercises: 14

Answer

$\frac{8}{3}$

Work Step by Step

$y=x^2$, $y=4x-x^2$ Find their intersections: $x^2=4x-x^2$ $2x^2-4x=0$ $x(2x-4)=0$ $x=0$ or $2x-4=0$ If $2x-4=0$, then $2x=4$, so $x=2$. So the intersections are $x=0$ and $x=2$. Note that between 0 and 2 $4x-x^2>x^2$, so the integrand is $(4x-x^2)-x^2$. The area is: $\int_0^2((4x-x^2)-x^2)dx$ $=\int_0^2(4x-2x^2)dx$ $=(\frac{4x^2}{2}-\frac{2x^3}{3})|_0^2$ $=(\frac{4*2^2}{2}-\frac{2*2^3}{3})-(\frac{4*0^2}{2}-\frac{2*0^3}{3})$ $=(8-\frac{16}{3})-(0-0)$ $=\frac{8}{3}$
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