(∫_2^(-u/2+25/6))v\[DifferentialD]v

(1/5)(∫_1^2)((481/72)-((25 u)/12)+((u^2)/8))\[DifferentialD]u

x coordinate: It looks like x=u will work.  
y coordinate: Let’s assume y=a u+b v +c. We’d like (x,y)=(0,0) to correspond to (u,v)=(0,0).  That is:

({{({{y=a u+b v +c
0=a (0)+b (0) +c}})}})

That is, c=0.

We’d like (x,y)=(0,1) to correspond to (u,v)=(0,1).This implies b=1. 

Finally if (x,y)=(1,3) corresponds to (u,v)=(1,1), then a=2, so