Simplification of an expression of a double integral

Question

Is the following expression able to be simplified? $$I = \int_0^{x/2} \int_0^s f(s,r)\,dr\,ds + \int_{x/2}^x \int_{2s - x}^s f(s,r)\,dr\,ds . $$ Here $f(s,r)=u(r,2s+x-r)$ so that we could also write $$I = \int_0^{x/2} \int_0^s u(r,2s+x-r)\,dr\,ds + \int_{x/2}^x \int_{2s - x}^s u(r,2s+x-r)\,dr\,ds . $$ $u$ is unknown. I made a lot a variables substitutions but without any success. I feel that I can simplify it much more than that. Any ideas? Thanks.

Answer 1

With $s$ as the horizontal axis and $r$ as the vertical axis, the region of integration is as in the diagram:

So reversing the order of integration allows the integral to be written as the single double integral

$$ I=\int_0^x\int_r^{(r+x)/2}f(s,r)\,dsdr $$