Answer
$1$
Work Step by Step
We have: $Area=\int_{0}^{\infty} e^{-x} dx=\lim\limits_{a \to \infty}\int_{0}^{a} e^{-x} dx$
or, $Area=\lim\limits_{a \to \infty}[-e^{-x}]_{0}^a=\lim\limits_{a \to \infty}[-e^{-a}-(-e^{-1})]$
or, $Area=0-(-1)=0+1=1$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 8: Techniques of Integration - Section 8.8 - Improper Integrals - Exercises 8.8 - Page 501: 67
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