Give a convincing argument that f(x)dx=1

We start with the definition of the normal probability density function:

$f(x)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-(x-\mu {)}^2/2{\sigma }^2}$

where $\mu$ is the mean of the distribution and σ2 is the variance.

We then substitute this function into the integral:

${\int }_{-b}^{b}f(x)dx={\int }_{-b}^b\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-(x-\mu {)}^2/2{\sigma }^2}dx$

We can now use the following change of variables:

$z=\frac{x-\mu }{\sigma }$

This change of variables transforms the integral into the following:

${\int }_{-b}^{b}f(x)dx={\int }_{-(b-\mu )/\sigma }^{(b-\mu )/\sigma }\frac{1}{\sqrt{2\pi }}{e}^{-z^2/2}dz$