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: 56


The area of the shaded region is $22/15$.

Work Step by Step

The shaded region is bounded above by the curve $y=x^2$ and below by the curve $y=-2x^4$ and runs from $x=-1$ to $x=1$. Therefore, the area of the shaded region is $$A=\int^{1}_{-1}(x^2+2x^4)dx$$ $$A=\Big(\frac{x^3}{3}+\frac{2x^5}{5}\Big)\Big]^{1}_{-1}$$ $$A=\Big(\frac{1}{3}+\frac{2}{5}\Big)-\Big(-\frac{1}{3}-\frac{2}{5}\Big)$$ $$A=2\Big(\frac{1}{3}+\frac{2}{5}\Big)$$ $$A=2\Big(\frac{11}{15}\Big)=\frac{22}{15}$$
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