Prove that $ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$ 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.
 A: You can use Taylor's theorem with error.  You are on the right track.
Suppose we are creating a Taylor polynomial about $a$.  There is a theorem that states that if, for fixed numbers $$m \le f^{n+1}(t) \le M$$ for all $t$ on some interval containing $a$, then the error $E(x)$ satisfies the inequalities $$\frac{m(x-a)^n}{(n+1)!} \le E(x) \le \frac{M(x-a)^n}{(n+1)!}$$ for $x>a$. (This is theorem 7.7 in Apostol's Calculus Vol. 1)
For our problem, we are using a Taylor polynomial centered around $0$, so $a=0$.  Take the interval $[0,1]$.  Then, we have the bounds $0 \le f^{2n}(t) \le \frac{(2n)!}{n!}$.
Substituting $2n$ for $n+1$ and $2n-1$ for $n$, we get $$0 \le E(x) \le \frac{x^{2n-1}}{n!}$$
Integrating the error, we get $$0 \le \int_0^1 E(x) dx \le \frac{1}{n!\cdot 2n}$$
It is important to note that $n$ is $(d-1) / 2$ where $d$ denotes the degree of the polynomial, because we substituted $2n$ instead of $n+1$.
A: 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
$$
1.4625300625\le\int_0^1e^{x^2}\,\mathrm{d}x\le1.4626563252
$$
A: 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
$$  \int _{0}^{1}\!{{\rm e}^{{x}^{2}}}{dx} = e-\int _{0}^{1}\!2\,{x}^{2}{
{\rm e}^{{x}^{2}}}{dx} \leq 1.462863173 < 1.463 $$
