Answer
$$ - \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^1 {\left( {x - 2y} \right)} dxdy} \cr
& \int_0^1 {\left[ {\int_0^1 {\left( {x - 2y} \right)} dx} \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^1 {\left( {x - 2y} \right)} dx = \left[ {\frac{{{x^2}}}{2} - 2xy} \right]_{x = 0}^{x = 1} \cr
& = \left[ {\left( {\frac{{{{\left( 1 \right)}^2}}}{2} - 2\left( 1 \right)y} \right) - \left( {\frac{{{{\left( 0 \right)}^2}}}{2} - 2\left( 0 \right)y} \right)} \right] \cr
& = \frac{1}{2} - 2y \cr
& {\text{Therefore,}} \cr
& \int_0^1 {\int_0^1 {\left( {x - 2y} \right)} dxdy} = \int_0^1 {\left( {\frac{1}{2} - 2y} \right)dy} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\frac{1}{2}y - {y^2}} \right]_0^1 \cr
& = \left[ {\frac{1}{2}\left( 1 \right) - {{\left( 1 \right)}^2}} \right] - \left[ {\frac{1}{2}\left( 0 \right) - {{\left( 0 \right)}^2}} \right] \cr
& = - \frac{1}{2} \cr} $$
