Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 304: 44

Answer

$\displaystyle \frac{9}{2}$

Work Step by Step

Graph both functions in the same window (see below). On the interval $x\in[0,3]$, the graph of $y= x$ is above the graph of $y=x^{2}-2x.$ The area between two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$. Thus, $A=\displaystyle \int_{0}^{3}[x-(x^{2}-2x)]dx=\int_{0}^{3}(3x-x^{2})dx$ $=3[\displaystyle \frac{x^{2}}{2}]_{0}^{3}-[\frac{x^{3}}{3}]_{0}^{3}$ $=\displaystyle \frac{3}{2}(9-0)-\frac{1}{3}(27-0)$ $=\displaystyle \frac{27}{2}-9$ $=\displaystyle \frac{27-18}{2}$ $=\displaystyle \frac{9}{2}$
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