## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 110

#### 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 $\ln B^x=x\ln B$.

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