Answer
$$\int_{0}^{1} \int_{2 x}^{2}(x-y) d y d x=-1$$
Work Step by Step
Given $$\int_{0}^{1} \int_{2 x}^{2}(x-y) d y d x$$
So, we have
\begin{aligned}I&=\int_{0}^{1} \int_{2 x}^{2}(x-y) d y d x\\
&=\int_{0}^{1}\left[x y-\frac{y^{2}}{2}\right]_{2 x}^{2} d x \\&=\int_{0}^{1}\left[x(2)-\frac{2^{2}}{2}\right]-\left[x(2 x)-\frac{(2 x)^{2}}{2}\right] d x\\
&=\int_{0}^{1}[2 x-2]-\left[2 x^{2}-2 x^{2}\right] d x\\
& =\int_{0}^{1} (2 x-2 )d x\\
&=\left[x^{2}-2 x\right]_{0}^{1}\\
&=1-2\\
&=-1
\end{aligned}
