Prove that $ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$

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 $n+1$.

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 $$ 1.4625300625\le\int_0^1e^{x^2}\,\mathrm{d}x\le1.4626563252 $$