# Explain why $\int_0^{\infty} e^{-t}dt = 1$.

I would appreciate it if someone could show the reasoning for this very elementary result: $$\int_0^\infty e^{-t}dt = 1$$

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$$\int e^{-t} dt = \int d(-e^{-t}) = -e^{-t} + c$$ Hence, $$\int_a^b e^{-t} dt = -e^{-b} + e^{-a}$$ $$\int_0^{\infty} e^{-t} dt = \lim_{b \to \infty} \int_0^b e^{-t} dt = \lim_{b \to \infty} \left(-e^{-b} + e^0 \right) = 1 - \lim_{b \to \infty} e^{-b} = 1 - 0 = 1$$

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Rats! When I began writing my question only the first line in your post appeared...Oh, well, +1. –  DonAntonio Feb 3 '13 at 4:30
I don't think anyone can give a better answer to a beginner then that. Just a straightforward calculation with the precise definition of limits. Couldn't be easier. –  Mathemagician1234 Feb 3 '13 at 4:36
Thanks! I was forgetting about the $e^0$. :-) –  Larry Freeman Feb 3 '13 at 4:37