University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 58

Answer

$4$

Work Step by Step

We can calculate the area, by integrating the function from -2 to 2. $\int_{-2}^{2} (3x^2-3)dx =[x^3-3x]_{-2}^{2}=[2^3-3\cdot2]-[(-2)^3-3\cdot(-2)]=[8-6]-[-8+6]=2-(-2)=4$
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