Prove $1.43 < \int_0^1 e^{x^2}\,\mathrm{d}x < \frac{e+1}{2}$

Prove $$1.43<\int_0^1 e^{x^2}\,\mathrm{d}x<\frac{e+1}{2}$$

What I did:

As I have no idea how to approach the left inequality I work with $$\int_0^1 e^{x^2} \mathrm{d}x<\frac{e+1}{2} \iff \int_0^1 e^{x^2} \mathrm{d}x<\int_0^1 \frac12 (e+x)\mathrm{d}x \iff e^{x^2}<\frac12(e+x) \iff x^2<\log (e+x)-\log 2$$

I don't know how to proceed.

Hint: You should be able to get the inequality $1.43\lt\int_0^1 e^{x^2}dx$ from

$$e^{x^2}\gt1+x^2+{1\over2}x^4+{1\over6}x^6+{1\over24}x^8+\cdots+{1\over n!}x^{2n}$$

for some truncation $n$.

Added later: Here's a way to get the upper bound. Using the fact that $x^2\le x$ for $0\le x\le1$ and $e\lt3$, one has

$$\int_0^1e^{x^2}\,dx\le\int_0^1e^x\,dx=e-1\lt{e+1\over2}$$

We have: $$\int_{0}^{1}e^{x^2}\,dx = e\cdot\int_{0}^{1}e^{1-x^2}\,dx = e\cdot\int_{0}^{1}e^{x(2-x)}\,dx=2e\cdot\int_{0}^{1/2}e^{4x(1-x)}\,dx$$ and since: $$ e^z = \sum_{j=0}^{+\infty}\frac{z^j}{j!} $$ we have: $$\begin{eqnarray*}\int_{0}^{1}e^{x^2}\,dx &=& 2e\cdot\int_{0}^{1/2}\sum_{j=0}^{+\infty}\frac{(-1)^j(4x(1-x))^j}{j!}\,dx =2e\cdot\sum_{j=0}^{+\infty}\frac{(-4)^j B(1/2,j+1,j+1)}{j!}\\&=&\frac{e}{2}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}4^j}{j!\binom{2j}{j}}=\frac{e\sqrt{\pi}}{2}\sum_{j=0}^{+\infty}\frac{(-1)^j}{\Gamma(j+3/2)}.\end{eqnarray*}$$ (see this question about an identity for the incomplete Beta function)

By summing the terms up to $j=5$ in the last series we get: $$1.469\ldots = \frac{73}{135}e\geq \int_{0}^{1} e^{x^2}\,dx \geq \frac{207}{385}e = 1.461\ldots. $$