Prove $\iint_D f(x + y) dA = \int_0^1 uf(u)du$ for $D = \{ (x, y) \in \mathbb{R}, 0 \leq x+y \leq 1, x \ge 0, y \ge 0\} $ and $f$ is a continuous function over $D$.
I have only gotten that $$du = \sqrt{x^2 + y^2} \text{d}xsin(\pi/4) = \sqrt{x^2 + y^2}\text{d}ysin(\pi/4) = y\text{d}x = x\text{d}y$$ drawing a picture of the region of integration and seeing what region would correspond to a change $\text{d}u$.
The region between the axes and the blue lines is what I thought would be $\Delta u$.
Your double integral is given by $$\int_0^1 \int_0^{1-x} f(x+y) \, dy \, dx.$$ Change variables in the inner integral: $u = x+y$, $du = dy$, to get $$\int_0^1 \int_x^1 f(u) \, du \, dx.$$ Change the order of integration: $$\int_0^1 \int_0^u f(u) \, dx du.$$ Finally observe $$\int_0^u f(u)\, dx = uf(u).$$