How to approximate the integral of the standard normal distribution. So I have this eqn.
$$
f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}}
$$
I need to find:
$$
\int\limits_{-1}^1 f(x)dx
$$
So I want to use this series to integrate. I know that:
$$
e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!}
$$
So without considering the coefficient of the function, I can build my eqn into this series as such:
$$
e^{\frac{-x^2}{2}} =\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!}
$$
This is where I get lost. What do I do with the coefficient of e. How do I build this into a series and include the coefficient of e:
$$
\frac {1}{\sqrt{2\pi}}\
$$
 A: Ok I figured it out.
$$
f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}}
$$
I need to find:
$$
\int\limits_{-1}^1 f(x)dx
$$
So I want to use this series to integrate. I know that:
$$
e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!}
$$
So without considering the coefficient of the function, I can build my eqn into this series as such:
$$
 e^{\frac{-x^2}{2}}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^nn!}
$$
So we can just build up and get the origial.
$$
\frac {1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}=\frac {1}{\sqrt{2\pi}} \sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!}
$$
Then all we need to do is expand out the series to a few values of n, then we can anti-differentiate the series, and then we can solve using the FTC, where the integral from [a,b] = F(b) - F(a).
A: Hint: You cannot find an explicit answer for your integration. However the integration of normal distribution can be stated  in terms of Q Function. 
$$
Q(x)=\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-t^2}{2}}dt
$$
There are various estimates for this function which may be useful. For instance for $x>0$:
$$
\frac{x}{(x^2+1)\sqrt{2\pi}}e^{\frac{-x^2}{2}} \leq Q(t)\leq \frac{1}{x\sqrt{2\pi}}e^{\frac{-x^2}{2}}
$$
Remark: If you want to continue working with series, it is enough to take the coefficient of the exponential function out of the integration. 
