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berljivx8

Answered question

2021-12-10

Need to find the Laplace transform of this:

f (t) = e^{- 2 t}

Answer & Explanation

Timothy Wolff

Beginner2021-12-11Added 26 answers

By the substitution

u = (2 + s) t w e \geq t \int_{0}^{\infty} e^{- (2 + s) t} d t = \frac{1}{2 + s} \int_{0}^{\infty} e^{- u} d u = \frac{1}{2 + s}

servidopolisxv

Beginner2021-12-12Added 27 answers

There is no need of IBP.

\int_{0}^{\infty} f (t) e^{- s t} d t = \int_{0}^{\infty} e^{- 2 t} e^{- s t} d t = \int_{0}^{\infty} e^{- (s + 2) t} d t = \frac{- 1}{s + 2} [0 - 1] = \frac{1}{s + 2}

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