How to evaluate $\int_0^1 \mathrm e^{-x^2} \,\mathrm dx$ using power series? I'm trying to evaluate 
$$\int_0^1 \mathrm e^{-x^2} \, \mathrm dx$$ 
using power series. 
I know I can substitute $x^2$ for $x$ in the power series for $\mathrm e^x$:
$$1-x^2+ \frac{x^4}{2}-\frac{x^6}{6}+ \cdots$$ 
and when I calculate the antiderivative of this I get 
$$x-\frac{x^3}{3}+ \frac{x^5}{5\cdot2}-\frac{x^7}{7\cdot6}+ \cdots$$ 
How do I evaluate this from $0$ to $1$?

$$\left[ x-\frac{x^3}{3}+ \frac{x^5}{5*2}-\frac{x^7}{7*6}+ \dots \right]_0^1 = \left( 1-\frac{1}{3} + \frac{1}{5 \cdot 2} - \frac{1}{7 \cdot 6} + \dots \right) - 0 = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)n!}$$ 
So  $\int_{0}^{1}e^{-x^2}=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)n!}$, that is, the answer is whatever the series converges to.
