I've already posted in the econ.stack, since the model I’m having trouble with is an econ one (link). However, I feel the nature of my doubt is more about math than the actual econ model behind it, so I decided to post also here changing my question accordingly.
The author identifies two product characteristics (x,y) that are orthogonal and are uniformly distributed over the unit square [0,1]x[0,1].
I understand the mechanics of the model, but I have a problem with the math behind some profit functions and consumer surplus functions. In particular, the integrals over the pdf/cdf really bug me.
Take eq. (3) for example:
$$CS^{fp} = 2 \int_0^{1-x} \int_0^{1} (v-t-t(x+y)) f(x) f(y) dx dy = v - \frac{5t}{3} $$
given $f(x)$ is the std. uniform I know that this density over the 0-1 interval is just 1 and I can basically ignore it, correct? An alternative view should be to consider $\int f(x)f(y) dx = \int f(x,y) dx = f(y)$ given independence. Correct?
If so, I can write
$$CS^{fp} = 2 \int_0^{1-x} f(y) \left[ v -t -t \int_0^{1} (x+y) dx \right]dy =$$
$$CS^{fp} = 2 \int_0^{1-x} f(y) \left[ v -ty - \frac32 t \right]dy =$$
If this is correct what is then $f(y)$ in this interval? How do I proceed?
My guess is... $$CS^{fp} = 2 \int_0^{1-x} v -ty - \frac32 t dy = 2\left[ vy - \frac{t y^2}{2} - \frac34 ty^2 \right]_0^{1-x} = 2\left[ vy - ty^2\frac{5}{4} \right]_0^{1-x}$$
$$ = 2v(1-x) - \frac52 t (1-x)^2 = $$
Yes, being $f(x)=f(y)=1$ you can write them off.
Your integral becomes, differently written:
$$2\int_0^1\left[\int_0^{1-x}(v-t-tx-ty)dy \right]dx=$$
$$=2\int_0^1\left[vy-ty-txy-\frac{t}{2}y^2\right]_{y=0}^{y=1-x}dx=\dots=v-\frac{5}{3}t$$