\begin{gather*} \iint\limits _{T}f(x,y)\;\mathrm{d}x\mathrm{d}y=\int\limits _{y=0}^{2}\left(\int\limits _{x=0}^{1}(4x-3y)^{3}\;\mathrm{d}x\right)\mathrm{d}y=\\ =\int\limits _{y=0}^{2}\left(\left[\dfrac{(4x-3y)^{4}}{16}\right]_{x=0}^{1}\right)\mathrm{d}y=\int\limits _{y=0}^{2}\left(\dfrac{(4\cdot1-3y)^{4}}{16}-\dfrac{(4\cdot0-3y)^{4}}{16}\right)\mathrm{d}y=\\ =\int\limits _{y=0}^{2}\left(\dfrac{(4-3y)^{4}}{16}-\dfrac{(-3y)^{4}}{16}\right)\mathrm{d}y=\int\limits_{y=0}^{2}\dfrac{(4-3y)^{4}}{16}\;\mathrm{d}y-\int\limits_{y=0}^{2}\dfrac{(-3y)^{4}}{16}\;\mathrm{d}y=\\ =\dfrac{1}{16}\cdot\left[\int\limits _{y=0}^{2}(4-3y)^{4}\;\mathrm{d}y-\int\limits _{y=0}^{2}(-3y)^{4}\;\mathrm{d}y\right]=\dfrac{1}{16}\cdot\left\{ \left[-\dfrac{(4-3y)^{5}}{15}\right]_{0}^{2}-\left[\dfrac{81}{5}y^{5}\right]_{0}^{2}\right\} =\\ =\dfrac{1}{16}\cdot\left\{ \left(\left[-\dfrac{(4-3\cdot2)^{5}}{15}\right]-\left[-\dfrac{(4-3\cdot0)^{5}}{15}\right]\right)-\left(\left[\dfrac{81}{5}\cdot2^{5}\right]-\left[\dfrac{81}{5}\cdot0^{5}\right]\right)\right\} =\\ =\dfrac{1}{16}\cdot\left[\left(\dfrac{32}{15}+\dfrac{1024}{15}\right)-\left(\dfrac{2592}{5}\right)\right]=\dfrac{1}{16}\cdot\left(\dfrac{352}{5}-\dfrac{2592}{5}\right)=-28\end{gather*}