Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 51



Work Step by Step

We have $$ \int_{1}^{3}[t] d t= \int_{1}^{2}[t] d t+\int_{2}^{3}[t] d t\\ = \int_{1}^{2}1 d t+\int_{2}^{3} 2d t\\ =t|_1^2+2t|_2^3=1+2=3. $$
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