MML Discourse archived in May, 2026

U-Subs and Z-Scores

mark

Use u-substitution to translate the normal integral

\frac{1}{\sqrt{8\pi}}\int_{-2}^5e^{-(x-3)^2/8}\,dx

into a standard normal integral.

User 003

The form of the normal integral is:

\frac{1}{\sqrt{2\pi}\sigma}\int_{a}^{b}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} dx

If we place the above normal integral in this form, we can see that \mu=3 and \sigma=2:

\frac{1}{2\sqrt{2\pi}}\int_{-2}^{5}e^{-\frac{1}{2}(\frac{x-3}{2})^2} dx

We also need to rearrange slightly to bring \frac{1}{\sigma} into the integral:

\frac{1}{\sqrt{2\pi}}\int_{-2}^{5}e^{-\frac{1}{2}(\frac{x-3}{2})^2} \frac{1}{2}dx

Now we can substitute in u=\frac{x-3}{2} and du=\frac{1}{2}dx as well as the new bounds determined by evaluating u for -2 and 5:

\frac{1}{\sqrt{2\pi}}\int_{-\frac{5}{2}}^{1}e^{-\frac{1}{2}u^2} du

This form is the standard normal integral.

User 009
\frac{1}{\sqrt{8\pi}}\int_{-2}^5e^{-\frac{(x-3)^2}{8}}dx\\ \\ \frac{1}{2\sqrt{2\pi}}\int_{-2}^5e^{-\frac{(x-3)^2}{8}}dx\\ \\ \frac{1}{2\sqrt{2\pi}}\int_{-2}^5e^{-\frac{1}{2}\frac{(x-3)^2}{4}}dx\\ \\ \frac{1}{2\sqrt{2\pi}}\int_{-2}^5e^{-\frac{1}{2}{u^2}}dx\\ \\ u^2 = \frac{(x-3)^2}{4}\\ \\ u = \frac{(x-3)}{2}\\ \\ x = 5, u = 1\\ x = -2, u = -5/2\\ \\ du = \frac{dx}{2}\\ 2du = dx\\ \\ \frac{1}{2\sqrt{2\pi}}\int_{\frac{-5}{2}}^1e^{-\frac{1}{2}{u^2}}2du\\ \\

2's cancel

\frac{1}{\sqrt{2\pi}}\int_{\frac{-5}{2}}^1e^{-\frac{1}{2}{u^2}}du\\