Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.1 Definition of the Laplace Transform - Problems - Page 675: 2

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

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}