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Use the given transformation to evaluate the integral.
$$\iint_{R}(x-3 y) d A$$ . where $$R$$ is the triangular region with
vertices $$(0,0),(2,1),$$ and $$(1,2) ; \quad x=2 u+v, y=u+2 v$$

$$x=2 u+v \quad , \quad[y=u+2 v) * 2$$

$$2 y=2 u+4 v$$

$$x=2 u+v$$
$$2 y=2 u+4 v$$
$$x-2 y=-3 v$$

$$v=\frac{1}{3}[2 y-x]$$

$$u=y-2 y$$
$$u=y-2 \frac{1}{3}[2 y-x]$$
$$u=-\frac{1}{3} y+\frac{2}{3} x$$

$$(1)(0,0) \Rightarrow(0,0)$$
$$v=\frac{1}{3}[0-0]=0$$
$$u=0$$
$$(2)(1,2) \Rightarrow(0,1)$$
$$v=\frac{1}{3}[2 \times 2-1]=1$$
$$u=\frac{-1}{3}\times 2+\frac{2}{3} \times 1=0$$

$$(3)(2,1)\Rightarrow(1,0)$$

$$\iint_{R\rightarrow(x, y)} f(x, y) d A=\iint_{S\rightarrow (u, v)} f\left[x(u, v), y(u, v)\right] | \frac{\partial(x, y)}{\partial(u, v)} | d v d u$$

$$\left|\begin{array}{ll}{\frac{\partial x}{\partial u}} & {\frac{\partial x}{\partial v}} \\ {\frac{\partial y}{\partial u}} & {\frac{\partial y}{\partial v}}\end{array}\right|=\left|\begin{array}{ll}{2} & {1} \\ {1} & {2}\end{array}\right|=2 * 2-1 * 1=4-1=3$$

$$\int_{R} \int(x-3 y) d x d y =\int_{0}^{1} \int_{0}^{1-u}(2 u+v)-3(u+2 v) * 3 d v d u$$

$$\int_{0}^{1} 3\left[-u v-5 \frac{(1-u)^{2}}{2}\right] d u=-3$$

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