- #1

quietrain

- 655

- 2

^{x}

the taylor expansion becomes 1+x+x

^{2}/2+...

integrating becomes x+x

^{2}/2+x

^{3}/6+...+c

so how do we know that c = 1? for it to become back to e

^{x}

becos it is said that integral of e

^{x}= e

^{x}

do we just let x be 0 to find c = 1? does it work for all expressions?

thanks a lot!