Question

Determine the Laplace transform of the following integrals. L\{\int_0^t e^{-2t}t^2 dt\}

Laplace transform
ANSWERED
asked 2021-09-03

Determine the Laplace transform of the following integrals
\(L\left\{\int_0^t e^{-2t}t^2 dt\right\}\)

Expert Answers (1)

2021-09-04

\(L\left\{\int_0^t e^{-2t}t^2 dt\right\}\)
We know that , if \(\displaystyle{L}{\left[{f{{\left({t}\right)}}}\right]}={F}{\left({s}\right)}\)
then \(\displaystyle{L}{\left[{\int_{{0}}^{{t}}}{f{{\left({v}\right)}}}{d}{v}\right]}={\frac{{{F}{\left({s}\right)}}}{{{s}}}}\)
Here we have
\(\displaystyle{\int_{{0}}^{{t}}}{e}^{{-{2}{t}}}{t}^{{2}}{\left.{d}{t}\right.}\)
so \(\displaystyle{f{{\left({t}\right)}}}={e}^{{-{2}{t}}}{t}^{{2}}\)
Since , if \(\displaystyle{L}{\left[{g{{\left({t}\right)}}}\right]}={G}{\left({s}\right)}\) then
\(\displaystyle{L}{\left[{e}^{{{a}{t}}}{g{{\left({t}\right)}}}\right]}={G}{\left({s}-{a}\right)}\)
so it we take \(\displaystyle{g{{\left({t}\right)}}}={t}^{{2}}\)
\(\displaystyle\Rightarrow{G}{\left({s}\right)}={L}{\left[{t}^{{2}}\right]}={\frac{{{2}}}{{{s}^{{3}}}}}\)
and hence
\(\displaystyle{L}{\left[{e}^{{-{2}{t}}}{t}^{{2}}\right]}={\left({\frac{{{2}}}{{{s}^{{3}}}}}\right)}_{{{s}\rightarrow{s}-{\left(-{2}\right)}}}\)
\(\displaystyle\Rightarrow{L}{\left[{e}^{{-{2}{t}}}{t}^{{2}}\right]}={\frac{{{2}}}{{{\left({s}+{2}\right)}^{{3}}}}}\)
Then
\(\displaystyle{L}{\left[{\int_{{0}}^{{t}}}{e}^{{-{2}{t}}}{t}^{{2}}{\left.{d}{t}\right.}\right]}={\frac{{{\frac{{{2}}}{{{\left({s}+{2}\right)}^{{3}}}}}}}{{{s}}}}\)
\(\displaystyle\Rightarrow{L}{\left[{\int_{{0}}^{{t}}}{e}^{{-{2}{t}}}{t}^{{2}}{\left.{d}{t}\right.}\right]}={\frac{{{2}}}{{{s}{\left({s}+{2}\right)}^{{3}}}}}\)

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