What substitution/manipulation was done here?

[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.

 

[PeroK]

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.

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

 

[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.

 

[theodoros.mihos]

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

 

[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.