Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 27



Work Step by Step

$\int^{4}_{-4}dx$ =$\int^{0}_{-4}|x|dx+\int^{4}_{0}|x|dx$ =-$\int^{0}_{-4}x dx+\int^{4}_{0}xdx$ =$[-\frac{x^2}{2}]_{-4}^0+[\frac{x^2}{2}]_0^4$ =$(\frac{0^2}{2}+\frac{(-4)^2}{2})+(\frac{4^2}{2}-\frac{0^2}{2})$ =16
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.