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 - Page 321: 58



Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.