Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 43



Work Step by Step

$\int_{0}^{2}(2t-3)dt$ =2$\int_{1}^{1}tdt -\int^{2}_{0}$ =2[$\frac{2^2}{2}-\frac{0^2}{2}$]-3(2-0) =4-6 =-2
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.