University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 97



Work Step by Step

$$A=\int^{-1}_{-2}e^{-(x+1)}dx$$ We set $u=-(x+1)$, which means $$du=-dx$$ $$dx=-du$$ For $x=-1$, $$u=-(-1+1)=0$$ For $x=-2$, $$u=-(-2+1)=1$$ Therefore, $$A=-\int^0_1e^udu$$ $$A=-e^u\Big]^0_1$$ $$A=-(e^0-e^1)$$ $$A=-(1-e)$$ $$A=e-1$$
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