I am looking at the derivation of the kinetic energy of a rigid body, with an initial velocity:
$v = v_{com} + w \times r$
I start with: $T = \int \frac12 v \cdot v \ dm=\int \frac12(v_{com} + w \times r) \ \cdot \ (v_{com} + w \times r) \ dm$
and after expanding all terms and substituting known variables I end up with
$T = \frac12M(v_{com} \cdot v_{com})+\frac12 (w \cdot Iw)+Mv_{com} \cdot w \times r $
I was wondering why this last term ends up disappearing since the final solution listed in my textbook is $T = \frac12M(v_{com} \cdot v_{com})+\frac12 (w \cdot Iw) $
When you expand $T = \int \frac12(v_{com} + \omega \times r) \ \cdot \ (v_{com} + \omega \times r) \ dm$, you get
$$T = \frac 12 \int v_{com}^2 \,dm + \frac 12 \int \|\omega \times r\|^2\, dm + \int v_{com}\omega \times r\,dm$$
As you've noted, the first two terms simplify to $\frac 12 Mv_{com}^2 + \frac 12(\omega \cdot I\omega)$. But in the last term, $r$ is not constant, so it does not simplify to $Mv_{com}\omega \times r$. And as I said in the comment, $r$ is not even defined outside the integral.
Instead, since $v_{com}$ and $\omega$ are constant, we get $$\int v_{com}\omega \times r\,dm = v_{com}\omega \times \int r\,dm$$
So the question remains, what is $\int r\,dm$? This can be a little tricky to recognize. $r$ is the location of the differential element $dm$, relative to the center of mass. If we fix an arbitrary coordinate system and let $R$ be the position vector, then $r = R - R_{com}$, where $R_{com}$ is the center of mass. So $$\int r\,dm = \int (R - R_{com})\,dm = \int R\,dm - MR_{com}$$
But the center of mass is defined by $$R_{com} := \frac{\int R\,dm}M$$
So you end up with $$\int r\,dm = 0$$