Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 1



Work Step by Step

The area $(A)$ can be calculated as: $A=\int_{-2}^2 [(2-x^2)-(-2)] \ dx \\= \int_{-2}^2 (4-x^2) \ dx \\=[4x-\dfrac{x^3}{3}]_{-2}^2 \\=\dfrac{16}{3}+\dfrac{16}{3} \\=\dfrac{32}{3}$
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.