Equation of line (0,0) to (3,-3) is y=-x.
So make x=t and y=-t and substitute in the integral !
for t going from 0 to 3 (since it's x)
Equation of line (3,-3) to (6,0) is y=x-6.
x=t and y=t-6
for t going from 3 to 6.
So this should give for the first one :
(note that since y=-t, dy=-dt)