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
$$ \int _{0}^{1}\!{{\rm e}^{{x}^{2}}}{dx} = e-\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx} \leq 1.462863173 < 1.463 $$