University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 65


The area of the region is $48/5$.

Work Step by Step

$y=x^4$ and $y=8x$ 1) Draft the graph The graph is enclosed below. From the graph, we choose integration with respect to $x$ to employ here for calculation of the region area. 2) Find the limits of integration: We can find the limits of integration by finding points of intersection between the curve and the line. $$x^4-8x=0$$ $$x(x^3-8)=0$$ $$x(x-2)(x^2+2x+4)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=2$$ So, the upper limit is $2$ and the lower one is $0$. 3) Find the area: Looking at the draft, we see that the region from $x=0$ to $x=2$ is bounded above by $y=8x$ and below by $y=x^4$. So, according to definition, the area of the region is $$A=\int^2_{0}[8x-x^4]dx$$ $$A=4x^2-\frac{x^5}{5}\Big]^2_{0}$$ $$A=4\times2^2-\frac{2^5}{5}$$ $$A=16-\frac{32}{5}=\frac{48}{5}$$
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