Hello,

Make a parametrization.

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)