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If $A(x)=\int_{-1}^{x}e^{-|t|}dt$, then what is $\lim_{x\to \infty}A(x)$?

I am not able to make any progress on this problem.I hope someone can help me with this.

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My first downvote! –  user54807 Jan 27 '13 at 5:23
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2 Answers 2

up vote 1 down vote accepted

$\int_{-1}^{x}e^{-|t|}dt$=$\int_{-1}^{0}e^{t}dt$+$\int_{0}^{x}e^{-t}dt$. This is because |t|=-t if t<0 and |t|=t if $t{>=}0$.
$\int_{-1}^{0}e^{t}dt$=1-1/e and $\int_{0}^{x}e^{-t}dt$=-$\int_{0}^{-x}e^{t}dt$=-($e^{-x}$-1)

Thus $\int_{-1}^{x}e^{-|t|}dt$=$=2-1/e-$$e^{-x}$

In the limit x->inf

A=2-1/e

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sorry for the duplicated answer, when I started to write it there were no answers! –  julian fernandez Jan 27 '13 at 4:38
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If $x\ge 0$ then $$ A(x)=\int_{-1}^x e^{-t}\,dt = \int_{-1}^0 e^t\,dt + \int_0^x e^{-t}\,dt. $$ Integrate, then find $\lim\limits_{x\to\infty}$.

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