Question

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

Laplace transform

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

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
$$\displaystyle{L}{\left[{\int_{{0}}^{{t}}}{e}^{{-{2}{t}}}{t}^{{2}}{\left.{d}{t}\right.}\right]}={\frac{{{\frac{{{2}}}{{{\left({s}+{2}\right)}^{{3}}}}}}}{{{s}}}}$$
$$\displaystyle\Rightarrow{L}{\left[{\int_{{0}}^{{t}}}{e}^{{-{2}{t}}}{t}^{{2}}{\left.{d}{t}\right.}\right]}={\frac{{{2}}}{{{s}{\left({s}+{2}\right)}^{{3}}}}}$$