 # determine the Laplace transform of f. f{{left({t}right)}}={3}{t}{e}^{ -{{t}}} Aneeka Hunt 2020-12-15 Answered
determine the Laplace transform of f.
$f\left(t\right)=3t{e}^{-t}$
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Step 1
To determine:
The Laplace transformation of the function $f\left(t\right)=3t{e}^{-t}$
Formula Used:
$F\left(s\right)=L\left(f\left(t\right)\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
Step 2
Explanation:
Let $f\left(t\right)=3t{e}^{-t}$
By using the formula,
$F\left(s\right)=L\left(3t{e}^{-t}\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}3t{e}^{-t}dt$
$=3\underset{b\to \mathrm{\infty }}{lim}{\int }_{0}^{b}{e}^{\left(-1-s\right)t}t{e}^{-t}dt$
$=3\underset{b\to \mathrm{\infty }}{lim}\left[{\left(t\frac{1}{-1-s}{e}^{\left(-1-s\right)t}\right)}_{0}^{b}-{\int }_{0}^{b}\frac{1}{-1-s}{e}^{\left(-1-s\right)t}dt\right]$
$=\frac{3}{-\left(s+1\right)}\underset{b\to \mathrm{\infty }}{lim}\left[\left(b{e}^{\left(-1-s\right)b}\right)-{\int }_{0}^{b}{e}^{\left(-1-s\right)t}dt\right]$
$=-\frac{3}{s+1}\underset{b\to \mathrm{\infty }}{lim}{\int }_{0}^{b}{e}^{\left(-1-s\right)t}dt$
$=\frac{-3}{s+1}\underset{b\to \mathrm{\infty }}{lim}{\left[\frac{1}{-\left(s+1\right)}{e}^{\left(-1-s\right)t}\right]}_{0}^{b}$
$=\frac{3}{{\left(s+1\right)}^{2}}\underset{b\to \mathrm{\infty }}{lim}\left[-{e}^{\left(-1-s\right)b}+1\right]$
$⇒F\left(s\right)=\frac{3}{{\left(s+1\right)}^{2}}$