Question

Evaluate:

${\int }_0^{\infty }{e}^{-x}dx$

Answer 1

${\int }_0^{\infty }{e}^{-x}dx=[\frac{e^{-x}}{-\left(1\right)}{]}_0^{\infty }=[-e^{-x}{]}_0^{\infty }=[-e^{-\infty }-(-e^{-0}\left)\right]$

$=0+e^0$

$=1$

$[\because \int e^x.dx=e^x]$