Here is how to find the upper bound, integration by parts gives
$ \int _{0}^{1}\!{{\rm e}^{{x}^{2}}}{dx}={{\rm e}}-\int _{0}^{1}\!2 \,{x}^{2}{{\rm e}^{{x}^{2}}}{dx}$
Using the fact that
$ 2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{10}+{\frac {1}{ 60}}\,{x}^{12}\leq 2\,{x}^{2}{{\rm e}^{{x}^{2}}}$
gives
$ \int _{0}^{1} (\!2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{ 10}+{\frac {1}{60}}\,{x}^{12}){dx}\leq \int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx}\,,$
since both functions are positive. Multiplying both sides of the above inequality by -1, yields,
$-\int _{0}^{1}(\!2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{ 10}+{\frac {1}{60}}\,{x}^{12}){dx}\geq -\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx}$
adding e to both sides of the last inequality gives
$ e-\int _{0}^{1}(\!2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{ 10}+{\frac {1}{60}}\,{x}^{12}){dx}\geq e-\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx}$
Evaluating the integral of the approximate power series gives the upper bound
$ \int _{0}^{1}\!{{\rm e}^{{x}^{2}}}{dx} = e-\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx} \leq 1.462863173 < 1.463 $