Definite Integral of $e^{\large x^2}$ I know there's no elementary antiderivative of $e^{\large x^2}$.
But what if there's a definite integral like 
$$\int_0^1e^{\large x^2}\ dx\ ?$$
I tried using basic definite integral property like $\displaystyle\int^a_0f(x)\ dx =\int^a_0f(a-x)\ dx $ but I could see no way out.

Using Maclaurin series of exponential function, we will obtain
$$
e^{\large x^2}=\sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}.
$$
Hence
$$
\begin{align}
\int_{x=0}^1\ e^{\large x^2}\ dx&=\int_{x=0}^1\ \sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}\ dx\\
&=\sum_{n=0}^\infty\int_{x=0}^1\ \frac{x^{\large 2n}}{n!}\ dx\\
&=\left.\sum_{n=0}^\infty\frac{x^{\large 2n+1}}{(2n+1)\ n!}\right|_{x=0}^1\\
&=\sum_{n=0}^\infty\frac{1}{(2n+1)\ n!}\\
&\approx 1.4626517459.
\end{align}
$$
