http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx Example $1$, problem (c):
$\int \int (6x^2 -40y)\, dA$, of triangle $A$ having vertices $(0,3), (5,3)$ and $(1,1)$.
On the “$-20(3-2x)^2$” section of the D$1$ integral (the left one), why does distributing that out result in a different answer than using $u=3-2x$, and $-0.5\, du= 1$, to make that section be $$-\dfrac{10}3(3-2x)^3$$ I finished integration through distributing the section and got a different result than the picture did, because the picture used $u$ substitution. They also used $u$ substitution in the same manner in the D$2$ integral of the problem, and I once again got a different answer by distributing instead of using $u$ substitution.
Ok, I think there are a lot of things you want to double check in your computations.
First, in the exercise the expression they have is $+20(3-2x)^2$.
Second, I think you mean $-\frac{1}{2} du = dx$.
Third, with that substitution you find \begin{equation} \int_0^1 20(3-2x)^2 dx = - \int_3^1 10 u^2 du = - \frac{10}{3} \left. u^3\right\vert_{3}^{1} = - \frac{10}{3}+90 = \frac{260}{3} \ . \end{equation}
This is the same result that you find by expanding and integrating directly \begin{equation} \int_0^1 20(3-2x)^2 dx = \int_0^1 ( 80 x^2 - 240 x +180 ) dx = \frac{80}{3}-120+180 = \frac{260}{3} \ . \end{equation}