Answer
$1$
Work Step by Step
We have
$$
\int_{1}^{3}[t] d t= \int_{1}^{2}[t] d t+\int_{2}^{3}[t] d t\\
= \int_{1}^{2}1 d t+\int_{2}^{3} 2d t\\
=t|_1^2+2t|_2^3=1+2=3.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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