#### Answer

$$=L[e^{2t}]= \frac{1}{s-2}$$

#### Work Step by Step

Given $$f(t)=e^{2t}$$
So, we get
\begin{aligned} L[f(t)] & =\int_{0}^{\infty} e^{-s t}f(t) \ d t\\
& =\int_{0}^{\infty} e^{-s t}e^{2t} d t \\
&=\int_{0}^{\infty} e^{-(s-2) t} d t\\
&= -\frac{1}{s-2}e^{-(s-2) t}| _{0}^{\infty} \\
&= 0-(-\frac{1}{s-2} ) \\
&= \frac{1}{s-2} \\
\end{aligned}