i understand all of it.Hi,
Since most of the work is already done for us here (all the integrals written out)
the only thing left to do is to find the new limits of integration. This is very
obvious if you look at the integrals for V2 and V3 and then try to figure
out the new limits of integration for V23.
If we integrate a function from -inf to t and then subtract from that the integral
of that same function integrated from -inf to t-1, that means we are left with
a much 'smaller' section to integrate over.
Geometrically, if we subtract a large area from a somewhat larger area, we are left
with a smaller area.
See it now?