Prove-that-0-e-t-dt-1-

Question Number 163619 by Zaynal last updated on 08/Jan/22

P r o v e t h a t;

\int_{- \infty}^{0} e^{- ∣ t ∣} d t = 1

Commented by alephzero last updated on 08/Jan/22

\int_{- \infty}^{0} e^{- ∣ t ∣} d t = 1

\lim_{q \to - \infty} (\int_{q}^{0} e^{- ∣ t ∣} d t) = 1

\int e^{- ∣ t ∣} d t =

e^{t} + C \Leftrightarrow t ⩽ 0

e^{- t} + C \Leftrightarrow t ⩾ 0

\Rightarrow \int_{q}^{0} e^{- ∣ t ∣} d t = e^{0} - e^{q} = 1 - e^{q} \Leftrightarrow q ⩽ 0

\Rightarrow \lim_{q \to - \infty} (\int_{q}^{0} e^{- ∣ t ∣} d t) =

= \lim_{q \to - \infty} (1 - e^{q}) =

= \lim_{q \to - \infty} (1) - \lim_{q \to - \infty} (e^{q})

\forall a {a \in R ∣ a^{- \infty} := 0}

\Rightarrow 1 - \lim_{q \to - \infty} (e^{q}) = 1 - 0 = 1

\Rightarrow \int_{- \infty}^{0} e^{- ∣ t ∣} d t = 1 Q . E . D .

Commented by Zaynal last updated on 09/Jan/22

thank you sir . perpect

Answered by MJS_new last updated on 08/Jan/22

- \infty < t ⩽ 0 \Rightarrow ∣ t ∣= - t

\underset{- \infty}{\int^{0}} e^{- ∣ t ∣} d t = \underset{- \infty}{\int^{0}} e^{t} d t = {[e^{t}]}_{- \infty}^{0} = 0 - (- 1) = 1

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