Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - Chapter Review - Review Exercises - Page 418: 66

Answer

$32$

Work Step by Step

\[\begin{align} & \text{From the graph we can define the area} \\ & A=\int_{0}^{2}{\left( 5-{{x}^{2}}-{{x}^{2}}+3 \right)}dx+\int_{2}^{4}{\left( {{x}^{2}}-3-5+{{x}^{2}} \right)dx} \\ & A=\int_{0}^{2}{\left( 8-2{{x}^{2}} \right)}dx+\int_{2}^{4}{\left( 2{{x}^{2}}-8 \right)dx} \\ & \text{Integrating} \\ & A=\left[ 8x-\frac{2}{3}{{x}^{3}} \right]_{0}^{2}+\left[ \frac{2}{3}{{x}^{3}}-8x \right]_{2}^{4} \\ & \text{Evaluating we obtain} \\ & A=\left[ 8\left( 2 \right)-\frac{2}{3}{{\left( 2 \right)}^{3}} \right]+\left[ \frac{2}{3}{{\left( 4 \right)}^{3}}-8\left( 4 \right) \right]-\left[ \frac{2}{3}{{\left( 2 \right)}^{3}}-8\left( 2 \right) \right] \\ & A=\frac{32}{3}+\frac{32}{3}+\frac{32}{3} \\ & A=32 \\ \end{align}\]
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