Consider the function f: \mathbb R^2 \to \mathbb R with
f(x,y) = A {\color{red}K} + B y and a constant \color{red}K.
Which value of \color{red}K ensures
\displaystyle \int \int_{{\color{orange}D}} f(x,y) dA = I where
\color{orange}D is the triangle below?
{\color{red}K} =
By definition
\displaystyle
\int \int_{{\color{orange}D}} f(x,y) \, dA =
\int \int_{{\color{orange}D}} (A {\color{red}K} + B y) \, dA =
\int \int_{{\color{orange}D}} A {\color{red}K} \, dA + \int \int_{{\color{orange}D}} B y \, dA.
The 1st summand
\displaystyle \int \int_{{\color{orange}D}} A {\color{red}K} \, dA =
A {\color{red}K} \int \int_{{\color{orange}D}} 1 \, dA can be determined geometrically,
since \displaystyle \int \int_{{\color{orange}D}} 1 \, dA = area of the triangle
{\color{orange}D} = X \cdot Y = Y*X.
Combined the integral has the value
\displaystyle \int \int_{{\color{orange}D}} A {\color{red}K} \, dA = A*X*Y {\color{red}K}.
We contemplate (geometrically), that the 2nd summand
\displaystyle \int \int_{{\color{orange}D}} B y \, dA must vanish.
It holds
\displaystyle
\int \int_{{\color{orange}D}} B y \, dA =
B \int_{0}^{X}
\int_{fractionReduce(Y,X,small=true) x - Y}^{-fractionReduce(Y,X,small=true) x +Y}
y \, dy dx.
For the interval boundaries of the inner integration one sees
fractionReduce(Y,X) x - Y = - \left(-fractionReduce(Y,X) x +Y\right),
i.e. one of the boundary is the negative of the other one.
Due to symmetry of the odd function y \mapsto y it follows
\displaystyle
\int_{fractionReduce(Y,X,small=true) x - Y}^{-fractionReduce(Y,X,small=true) x +Y} y \, dy =0.
Eventually
\displaystyle \int \int_{{\color{orange}D}} f(x,y) dA = I = A*X*Y {\color{red}K}
and
{\color{red}K} = K.