I want to find $\int p(x) dx$ where: $$p(x)= \frac{e^{\frac{-x^2}{2}}}{\sqrt{2\pi}}.$$
This function can not be manually computed so I am using Maclaurin series to find the antiderivative. I believe the Maclaurin series for this function would be: $$\frac{1}{\sqrt{2\pi}} \sum_{n=0}^\infty \frac{\frac{-x^2}{2}^n}{n!}.$$
The bounds of integration would be from $-1$ to $1$. How would I create a Maclaurin series for the antiderivative from this?
On
The best you can do is approximate the value of a definite (not indefinite integral, as the Maclaurin series will be useless for this, unless you are satisfied with some other power series and an added constant) using the series you correctly computed.
Using the fact that this is an alternating, you can get a precise bound on the remainder.
We can integrate the Maclaurin Series term by term as follows. $$2\int_{0}^1\frac{1}{\sqrt{2\pi}} \sum_{n=0}^\infty \frac{\big(\frac{-x^2}{2}\big)^n}{n!}$$ $$\sqrt{\frac{2}{\pi}} \sum_{n=0}^\infty \int_{0}^1\frac{(-1)^nx^{2n}}{2^nn!}$$ $$\sqrt{\frac{2}{\pi}} \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2^nn!(2n+1)}\Bigg|_{0}^1$$ $$\sqrt{\frac{2}{\pi}} \sum_{n=0}^\infty \frac{(-1)^n}{2^nn!(2n+1)} \approx 0.682689492$$ From this series, we get an approximation of the true value.