Find the laplace transform of the following function f(t)=tu_2(t)

Question

Find the laplace transform of the following function

f (t) = t u_{2} (t)

Answer 1

$Step 1$
$We apply defination of Laplace transformation.$
$Step 2$
$There u_{2} (t) = {\begin{cases} 0 & if t < 2 \\ 1 & if t > 2 \end{cases}$
$Now f (t) = t u_{2} (t) = {\begin{cases} 0 & if t < 2 \\ 1 & if t > 2 \end{cases}$
$L (f (t)) = \int_{0}^{\infty} f (t) e^{- s t} d t$
$= \int_{0}^{2} f (t) e^{- s t} d t + \int_{2}^{\infty} f (t) e^{- s t} d t$
$= 0 + \int_{2}^{\infty} f (t) e^{- s t} d t$
$= \int_{2}^{\infty} t e^{- s t} d t$
$= {[t \int e^{- s t} d t]}_{2}^{\infty} - \int_{2}^{\infty} [\frac{d}{d t} \cdot t \int e^{- s t} d t] d t$
$= {[- \frac{t}{s} e^{- s t}]}_{2}^{\infty} + \frac{1}{s} \int_{2}^{\infty} e^{- s t} d t$
$= \frac{2}{s} e^{- 2 s} - \frac{1}{s^{2}} {[e^{- s t}]}_{2}^{\infty}$
$= e^{- 2 s} (\frac{2}{s} + \frac{1}{s^{2}}) = \frac{e^{- 2 s}}{s^{2}} (2 s + 1)$
$So L (f (t)) = \frac{e^{- 2 s}}{s^{2}} (2 s + 1)$

Find the laplace transform of the following function f(t)=tu_2(t)

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