A Nice Integral (5)

$\begin{aligned} I & = \int_0^1 \color{#3D99F6}{e^{x^2}} dx & \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \int_0^1 \color{#3D99F6} {\sum_{n=0}^\infty \frac {x^{2n}}{n!}} dx \\ & = \sum_{n=0}^\infty \int_0^1 \frac {x^{2n}}{n!} dx \\ & = \sum_{n=0}^\infty \frac {x^{2n+1}}{n!(2n+1)} \bigg|_0^1 \\ & = \sum_{n=0}^\infty \frac 1{n!(2n+1)} \\ & = 1 + \frac 13 + \frac 1{10} + \frac 1{42} + \frac 1{216} + \color{#3D99F6}{\frac 1{1320} + ...} & \small \color{#3D99F6}{\text{Terms }< 0.0007575} \\ & \approx \boxed{1.462} \end{aligned}$

I am going to evaluate $\int_{0}^{1} e^{x^2} dx$ by using $\color{#D61F06}{Taylor's \ theorem}$ of the mean:

$\large e^x = 1 + x + \frac{ x^2}{2!}+ \frac{x^3}{3!}+ \frac{ x^4}{4!}+ \frac{x^5 e^\xi}{5!}$ ; where $(0< \xi <x)$

Then replacing $x$ by $\color{#3D99F6}{x^2}$ we obtain:

$\large e^{x^2} = 1 +\color{#3D99F6}{ x^2} + \frac{ \color{#3D99F6}{x^4}}{2!}+ \frac{\color{#3D99F6}{x^6}}{3!}+ \frac{ \color{#3D99F6}{x^8}}{4!}+ \frac{\color{#3D99F6}{x^{10}} e^\xi}{5!}$ ; where $(0< \xi <x)$

Integrating from $0$ to $1$ :

$\large\int_{0}^{1}= \int_{0}^{1} (1 + x^2 + \frac{ x^4}{2!}+ \frac{x^6}{3!}+ \frac{ x^8}{4!})dx+ E$ , where $\large E= \int_{0}^{1} \frac{x^{10} e^\xi}{5!} dx$ , $\large E$ denotes the maximum error.

$\large= 1 + \frac{1}{3} + \frac{1}{5.2!} + \frac{1}{7.3!}+ \frac{1}{9.4!} + E = 1.4618 + E$ ;

Now $\large |E|= |\int_{0}^{1} \frac{x^{10} e^\xi}{5!} dx| \le |\int_{0}^{1} \frac{x^{10} e^\xi}{5!}| dx \le e\int_{0}^{1} \frac{x^{10}}{5!} dx = \frac{e}{11.5!} < 0.0021$

Thus the maximum error is less than $0.0021$ , and so the value of the integral is $1.46$ accurate to two decimal places.