Chapter 5.6, Problem 389E

[T] The normal distribution in probability is given

by $p\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-{\left(x-\mu \right)}^2/\left(2{\sigma }^2\right)}$ , where

$\sigma$ is the standard

deviation and $\mu$ is the average. The standard normal distribution in probability, $p_s$ , corresponds to $\mu =0$ and $\sigma =1$ . Compute the left endpoint estimates $R_{10}$ and $R_{100}$ of ${\displaystyle {\int }_{-1}^1\frac{1}{\sqrt{2\pi }}{e}^{-x^{2/2}}}dx$