The Laplace transform of u(t - 2) is a) 1/s + 2 b) 1/s - 2

The Laplace transform of $$\displaystyle{u}{\left({t}-{2}\right)}$$ is
(a) $$\displaystyle\frac{{1}}{{s}}+{2}$$
(b) $$\displaystyle\frac{{1}}{{s}}-{2}$$
(c) $$\displaystyle{e}^{{2}}\frac{{s}}{{s}}{\left({d}\right)}\frac{{e}^{{−{2}{s}}}}{{s}}$$

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Bentley Leach
We know that:
$$\displaystyle{L}{\left({u}{\left({t}\right)}\right)}=\frac{{1}}{{s}}$$
By t shifting theorem
$$\displaystyle{L}{\left({u}{\left({t}-{2}\right)}\right)}={e}^{{-{2}{s}}}{L}{\left({u}{\left({t}\right)}\right)}=\frac{{{e}^{{-{2}{s}}}}}{{s}}$$
$$\displaystyle{\left({d}\right)}\frac{{{e}^{{-{2}{s}}}}}{{s}}$$