Can I find integral of $e^x$ from its Taylor series expansion.

Question

I tried to do evaluate integral of $e^x$ from series this way I'm stuck :- So where am I mistaken ???! My solution here (https://i.stack.imgur.com/hCp1R.jpg)

Answer 1

Maybe try it this way:

$\begin{align} \int_c^xe^tdt&=\int_c^x\left(1+t+\frac{t^2}{2}+\frac{t^3}{3!}+\dots\right)dt\\ &=\int_c^xdt+\int_c^xtdt+\int_c^x\frac{t^2}{2}dt+\int_c^x\frac{t^3}{3!}dt+\dots\\ &=(x-c)+\left(\frac{x^2}{2}-\frac{c^2}{2}\right)+\left(\frac{x^3}{3!}-\frac{c^3}{3!}\right)+\dots\\ &=\left(1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\dots\right)-\left(1+c+\frac{c^2}{2}+\frac{c^3}{3!}+\dots\right)\\ &=e^x-e^c \end{align}$

Answer 2

The $C$ in the answer is an arbitrary constant and there is no difference between having $-1+C$ and having $C$. It is just a question of renaming the constant.

Answer 3

Since the indefinite integral is determined up to a constant, we just need to set the constant $C'=C-1$.