Question-105598

Question Number 105598 by mohammad17 last updated on 30/Jul/20

Answered by Aziztisffola last updated on 30/Jul/20

L {e^{ax}} = \int_{0}^{\infty} e^{- sx} e^{ax} dx = \int_{0}^{\infty} e^{- sx + ax} dx

=∣ \frac{1}{a - s} e^{(a - s) x} ∣_{0}^{x \to \infty} = 0 - \frac{1}{a - s} = \frac{1}{s - a} (s > a)

Hence L {e^{ax}} = \frac{1}{s - a}