Originally Posted by
DivideBy0 $\displaystyle r$ is just a constant, isn't it?
Then simply solve both equations for $\displaystyle y$:
$\displaystyle y = 1-x$
$\displaystyle y=\frac{r}{x}$
The graphs intersect at $\displaystyle \lbrace{(x,y): \left(\frac{1+\sqrt{1-4r}}{2},\frac{1-\sqrt{1-4r}}{2}\right),\left(\frac{1-\sqrt{1-4r}}{2},\frac{1+\sqrt{1-4r}}{2}\right)\rbrace}$
So you need to evaluate:
$\displaystyle \int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} (1-x)\,dx-\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} \frac{r}{x}\,dx$