The Weirstrass Test for absolute uniform convergence of improper integrals states that if there is a function $M(x)$ such that $|f(x,y)| \leq M(x)$ for all $y$,$x$ in the domain of consideration, then if $\int_a^b M(x)dx < \infty$ we have that $\int_a^b f(x,y)dx$ converges absolutely/uniformly.
So if $f(x,y)\geq 0$ for all $x,y$ in the domain of interest then I can just choose $M(x) = f(x,y)$. Then if $\int_a^bf dx$ converges then I'm done right?
I feel that I'm missing something here since I haven't seen this stated anywhere.
Choosing $M(x) = f(x,y)$ doesn't make sense, because $f(x,y)$ depends on $x$ and $y$ but $M(x)$ may only depend on $x$.
You may certainly define $M(x) = \sup_y f(x,y)$, which (when it's finite, which doesn't happen all the time) does supply an upper bound suitable for the Weierstrass test.