Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 47

Answer

$= \frac{32}{3}$
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Work Step by Step

$\int^{0}_{-2} (4-x^{2}) dx + \int^{2}_{0} (4-x^{2}) dx$ $= [4x - \frac{x^{3}}{3} |^{0}_{-2}] + [4x - \frac{x^{3}}{3} |^{2}_{0}]$ $= [(4(0) - \frac{0^{3}}{3}) - (4(-2) - \frac{(-2)^{3}}{3})] + [(4(2) - \frac{2^{3}}{3}) - (4(0) - \frac{(0)^{3}}{3})]$ $= [(0) - (-8 - \frac{-8}{3})] + [(8 - \frac{8}{3}) - (0)]$ $= [- (-\frac{16}{3})] + [\frac{16}{3}]$ $= [\frac{16}{3}] + [\frac{16}{3}]$ $= [\frac{16}{3}] + [\frac{16}{3}]$ $= \frac{32}{3}$
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