Answer
$-1$
Work Step by Step
Using the fact that $\frac{u'}{u}=\ln u $, we have
$$\int_{-e^2}^{-e} \frac{1}{t}dt= \ \ln( |t|)|_{-e^2}^{-e} \\
=\ln e-\ln e^2\\
=\ln e-2\ln e=1-2=-1. $$
where used the properties $\ln e=1$ and $\ln B^x=x\ln B $.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 110
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