Kenya Leonard

2022-07-19

find $\int p\left(x\right)dx$ where:$p\left(x\right)=\frac{{e}^{\frac{-{x}^{2}}{2}}}{\sqrt{2\pi }}.$

Marisa Colon

Expert

We can integrate the Maclaurin Series term by term as follows.
$2{\int }_{0}^{1}\frac{1}{\sqrt{2\pi }}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(\frac{-{x}^{2}}{2}{\right)}^{n}}{n!}$
$\sqrt{\frac{2}{\pi }}\sum _{n=0}^{\mathrm{\infty }}{\int }_{0}^{1}\frac{\left(-1{\right)}^{n}{x}^{2n}}{{2}^{n}n!}$
$\sqrt{\frac{2}{\pi }}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}{x}^{2n+1}}{{2}^{n}n!\left(2n+1\right)}{|}_{0}^{1}$
$\sqrt{\frac{2}{\pi }}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{2}^{n}n!\left(2n+1\right)}\approx 0.682689492$
From this series, we get an approximation of the true value.

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