Answer
$-3$
Work Step by Step
$\int \int_R 2(x-y) \space dx \space dy =\int \int_G (-2v)|\dfrac{\sigma(x,y)}{\sigma(u,v)}| \space du \space dv$
or, $=\int \int_G \space (-2v) \space du \space dv$
or, $=\int^1_0 \int^{3-3v}_{-3v}(-2v) \space du \space dv$
or, $=\int^1_0 (-2v) \times (3-3v+3v) \space dv $
or, $=[-3v^2]^1_0$
or, $=-3$
