# What substitution/manipulation was done here?

1. Jun 9, 2015

### MathewsMD

I've attached the image with this post. It looks fairly simply to go from the first line to the second line and almost like a uv substitution, but I tried plugging it in and playing around with the integral and cannot get the exact answer. I really feel like something's just going over my head or that I'm missing a step. Regardless, any help would be great!

Note: mu is the mean.

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2. Jun 9, 2015

### PeroK

If $\mu$ is the mean, and $f$ is the pdf, then isn't $\mu = \int xf(x)dx$?

3. Jun 9, 2015

### mathman

Expand $(x-\mu)^2 = x^2-2x\mu+\mu^2$ to get three integrals and use the definition of $\mu$ to get the second term.

4. Jun 10, 2015

### theodoros.mihos

By infinity limits the integral is the same for each value of $\mu$. Are you post all informations?

Last edited: Jun 10, 2015
5. Jun 10, 2015

### mathman

$\mu$ is a constant. Therefore $\int_{-\infty}^{\infty}\mu^2f(x)dx=\mu^2\int_{-\infty}^{\infty}f(x)dx=\mu^2$.

Also $\int_{-\infty}^{\infty}\mu xf(x)dx=\mu\int_{-\infty}^{\infty}xf(x)dx=\mu^2$.