Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises - Page 374: 53

Answer

$$A = \frac{{256}}{5}$$

Work Step by Step

$$\eqalign{ & \left( i \right){\text{From the graph we can see that the net area of the region is:}} \cr & A = \int_{ - 2}^2 {\left( {16 - {x^4}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left[ {16x - \frac{1}{5}{x^5}} \right]_{ - 2}^2 \cr & {\text{Evaluating}} \cr & A = \left[ {16\left( 2 \right) - \frac{1}{5}{{\left( 2 \right)}^5}} \right] - \left[ {16\left( { - 2} \right) - \frac{1}{5}{{\left( { - 2} \right)}^5}} \right] \cr & A = \frac{{128}}{5} + \frac{{128}}{5} \cr & A = \frac{{256}}{5} \cr} $$
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