TITLE: Prove that $ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$
QUESTION [12 upvotes]: Prove the following integral inequality:
$$ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$$
This is a high school problem. So far i did manage to prove that the integral is bigger than $1.462$ by using Taylor expansion, namely:
$$1.462\le 1.4625=\int_0^1 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+\frac{x^{10}}{120}\le  \int_0^1 e^{{x}^{2}}$$
For the right bound i'm still looking for a way. However, i wonder if there is an elegant way to solve both sides.

REPLY [5 votes]: Integrating the power series for $e^{x^2}$, term by term, gives
$$
\int_0^1e^{x^2}\,\mathrm{d}x=\sum_{k=0}^\infty\frac1{(2k+1)k!}
$$
Since
$$
\begin{align}
\sum_{k=n+1}^\infty\frac1{(2k+1)k!}
&\le\frac1{(2n+3)(n+1)!}\left[1+\frac1{n+2}+\frac1{(n+2)^2}+\dots\right]\\
&=\frac1{(2n+3)(n+1)!}\frac{n+2}{n+1}\\
&\le\frac1{(2n+1)(n+1)!}
\end{align}
$$
If we use $n=5$ in
$$
\sum_{k=0}^n\frac1{(2k+1)k!}\le\int_0^1e^{x^2}\,\mathrm{d}x\le\frac1{(2n+1)(n+1)!}+\sum_{k=0}^n\frac1{(2k+1)k!}
$$
we get
$$
1.4625300625\le\int_0^1e^{x^2}\,\mathrm{d}x\le1.4626563252
$$