I would like to simplify this integral, and I think I can perform a change of variables. Here's the integral. $t, y$ and $z$ are all scalars:
$$ \int_{0}^{1}f(ty,tz)\cdot z dt $$
Note that we have a "$g(x)=tx$" and a "$g'(x)=x$'' that could be used as the basis for a change of variables. However, the $g(x)$ is applied to two arguments to $f$ (both $y$ and $z$). Perhaps I can do the following?
$$ \int_{0}^{1}f(ty,\underbrace{tz}_{g(x)})\cdot \underbrace{z}_{g'(x)} dt = \int_{g(0)}^{g(1)}f(y,v) dv $$ Most of the change of variables rules are all about one argument, so it's not clear what I can do.
Update: I think I should consider $g(t)=zt$, and thus $g'(t)=z$. Thank you to @KurtG for suggesting this. Now we can write $ty=\frac{g(t)y}{z}$. I'm not yet sure how this affects the answer, but am still trying.
Thank you to the amazing @KurtG. for providing (basically) the answer in the comments. Glory belongs to him! I tried to get him to post this so he could get credit, but he declined.
Here's the idea: $g(t)=zt$, and thus $g'(t)=z$. Now we can write: \begin{equation} \int_{0}^{1}f(ty,tz)z, dt = \int_{g(0)}^{g(1)}f(vy/z,v), dv = \int_{0}^{z}f(vy/z,v), dv \end{equation}
Now, one should be mindful that $z$ coudl equal zero, which would create a division-by-zero problem with the final term on the far RHS. However, @KurtG. points out that if $z=0$, then the far LHS evaluates to zero as well (through the product with $z$).
Some interesting test cases are $f(y,z)=y+z$ and $f(y,z)=y^2+zy+z^2$. It checks out numerically.
Let me say: I wrote this problem as the bare minimum case of a larger problem, when I had a $f$ that takes a vector as an argument. The idea above generalizes to:
\begin{equation} \int_{0}^{1}f(tx_{1},tx_{2},...,tx_{N},tz)z, dt= \int_{g(0)}^{g(1)}f(vx_{1}/z,vx_{2}/z,...,vx_{N}/z,v), dv = \int_{0}^{z}f(vx_{1}/z,vx_{2}/z,...,vx_{N}/z,v), dv \end{equation} Hope I can help others in the same situation.