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Question Number 128182 by BHOOPENDRA last updated on 05/Jan/21

Commented by BHOOPENDRA last updated on 05/Jan/21

$$findlaplacetransformation?$$

Answered by mathmax by abdo last updated on 05/Jan/21

$$2)\{\begin{array}{l}\mathrm{f}\left(\mathrm{t}\right)={(\mathrm{t}-2)}^2\mathrm{if}\mathrm{t}>2\\ \mathrm{o}\mathrm{if}\mathrm{t}<2\Rightarrow \mathrm{L}\left(\mathrm{f}\left(\mathrm{t}\right)\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{tx}}{(\mathrm{x}-2)}^2\mathrm{dx}\end{array}$$$$={\int }_2^{+\mathrm{\infty }}{(\mathrm{x}-2)}^2{\mathrm{e}}^{-\mathrm{tx}}\mathrm{dx}={[-\frac{1}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{tx}}{(\mathrm{x}-2)}^2]}_{\mathrm{x}=2}^{\mathrm{\infty }}+2{\int }_2^{+\mathrm{\infty }}\frac{1}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{tx}}(\mathrm{x}-2)\mathrm{dx}$$$$=\frac{2}{\mathrm{t}}{\int }_2^{+\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{tx}}(\mathrm{x}-2)\mathrm{dx}$$$$=\frac{2}{\mathrm{t}}\{{[-\frac{1}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{tx}}(\mathrm{x}-2)]}_2^{\mathrm{\infty }}+{\int }_2^{\mathrm{\infty }}\frac{1}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{tx}}\mathrm{dx}\}$$$$=\frac{2}{{\mathrm{t}}^2}{\int }_2^{+\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{tx}}\mathrm{dx}=\frac{2}{{\mathrm{t}}^2}{[-\frac{1}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{tx}}]}_2^{\mathrm{\infty }}$$$$=\frac{2}{{\mathrm{t}}^2}\left(\frac{{\mathrm{e}}^{-2\mathrm{t}}}{\mathrm{t}}\right)=\frac{{2\mathrm{e}}^{-2\mathrm{t}}}{{\mathrm{t}}^3}$$$$$$

Commented by BHOOPENDRA last updated on 06/Jan/21

$$questionno.thirdsir?tqforthesecondone$$

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