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: 43

Answer

$\displaystyle \frac{48}{5}$

Work Step by Step

Graph both functions in the same window (see below). On the interval $x\in[0,2]$, the graph of $y= 8x$ is above the graph of $y=x^{4}.$ 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, $A=\displaystyle \int_{0}^{2}[8x-(x^{4})]dx=$ $=8[\displaystyle \frac{x^{2}}{2}]_{0}^{2}-[\frac{x^{5}}{5}]_{0}^{2}$ $=4(4-0)-\displaystyle \frac{1}{5}(32-0)$ $=16-\displaystyle \frac{32}{5}$ $=\displaystyle \frac{80-32}{5}$ $=\displaystyle \frac{48}{5}$
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