Answer
-2
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
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
