1. **Problem Statement:** Calculate the double integral $$\iint_R f(x,y) \, dA$$ where $$R = \{(x,y): 0 \leq x \leq 4, 0 \leq y \leq 2\}$$ and the function $$f(x,y)$$ is defined as: $$f(x,y) = \begin{cases} 2 & \text{if } 1 \leq x < 3, 0 \leq y \leq 2 \\ 3 & \text{if } 3 \leq x \leq 4, 0 \leq y \leq 2 \end{cases}$$ 2. **Formula and Explanation:** The double integral over a region $$R$$ is calculated by integrating the function over $$x$$ and $$y$$: $$\iint_R f(x,y) \, dA = \int_{x=a}^{b} \int_{y=c}^{d} f(x,y) \, dy \, dx$$ Since $$f(x,y)$$ is piecewise, we split the integral into two parts: - For $$1 \leq x < 3$$, $$f(x,y) = 2$$ - For $$3 \leq x \leq 4$$, $$f(x,y) = 3$$ 3. **Calculate each part:** - For $$x$$ in $$[0,1)$$, $$f(x,y) = 0$$ (not given, assume zero or no contribution) - For $$x$$ in $$[1,3)$$: $$\int_1^3 \int_0^2 2 \, dy \, dx = \int_1^3 \left[2y\right]_0^2 \, dx = \int_1^3 4 \, dx = 4(x)\big|_1^3 = 4(3-1) = 8$$ - For $$x$$ in $$[3,4]$$: $$\int_3^4 \int_0^2 3 \, dy \, dx = \int_3^4 \left[3y\right]_0^2 \, dx = \int_3^4 6 \, dx = 6(x)\big|_3^4 = 6(4-3) = 6$$ - For $$x$$ in $$[0,1)$$, since $$f$$ is not defined, assume zero contribution. 4. **Sum the parts:** $$8 + 6 = 14$$ 5. **Final answer:** $$\boxed{14}$$