what is integral of
$\displaystyle e^{-x^2}$ does not have an elementry antiderivative. The integral can be calulated using Taylor series.
$\displaystyle e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} $
So we can integrate this term by term to get
$\displaystyle \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!} $
This can be used to approximate the value to any accuracy you wish. The error bound is simple because the series is alternating.
You have already been told that this is not expressible a closed form in terms of elementary functions. You have been given a series approximation which will give the numerical value to any precision you require.
$\displaystyle \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}\\ \\ \phantom{SSSSSSS}=\sum_{n=0}^{N-1}\frac{(-1)^n}{(2n+1)n!} +R_N$
Where $\displaystyle |R_N|<\frac{1}{(2N+1)N!} $
CB
You have been told twice now, by TheEmptySet and Captain Black, that [itex]e^{-x^2}[/tex] does not have an "elementary" anti-derivative.
If you insist upon such an answer then it is $\displaystyle \frac{\sqrt{\pi}}{2} erf(1)$ where "erf" is the Gauss error function. It is defined as $\displaystyle \frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx$. Does that help?
If you multiply it by $\displaystyle 2/\sqrt\pi$, this integral becomes the error function, for which detailed tables of values are available (for example on the Wikipedia page that I just linked to). This gives erf(1) = 0.8427008. So your integral is that number times $\displaystyle \sqrt\pi/2$, or approximately 0.7468241.