We are asked to evaluate the double integral $\iint_R f(x, y) dA$, where $R = \{(x, y): 1 \le x \le 4, 0 \le y \le 2\}$ and $f(x, y)$ is a piecewise function. We will solve the first case: $f(x, y) = \begin{cases} 2 & 1 \le x < 3, 0 \le y \le 2 \\ 3 & 3 \le x \le 4, 0 \le y \le 2 \end{cases}$

1. Problem Description

We are asked to evaluate the double integral $\iint_R f(x, y) dA$, where $R = \{(x, y): 1 \le x \le 4, 0 \le y \le 2\}$ and $f(x, y)$ is a piecewise function. We will solve the first case:

$f(x, y) = \begin{cases} 2 & 1 \le x < 3, 0 \le y \le 2 \\ 3 & 3 \le x \le 4, 0 \le y \le 2 \end{cases}$

2. Solution Steps

The region $R$ is a rectangle defined by $1 \le x \le 4$ and $0 \le y \le 2$. Since $f(x,y)$ is defined piecewise based on the value of $x$, we split the region of integration into two parts:

$R_1 = \{(x, y): 1 \le x < 3, 0 \le y \le 2\}$

$R_2 = \{(x, y): 3 \le x \le 4, 0 \le y \le 2\}$

Then we can express the double integral as a sum of two double integrals:

$\iint_R f(x, y) dA = \iint_{R_1} f(x, y) dA + \iint_{R_2} f(x, y) dA$

Since $f(x, y) = 2$ on $R_1$ and $f(x, y) = 3$ on $R_2$, we have:

$\iint_{R_1} f(x, y) dA = \iint_{R_1} 2 dA = 2 \iint_{R_1} dA = 2 \cdot \text{Area}(R_1)$

$\iint_{R_2} f(x, y) dA = \iint_{R_2} 3 dA = 3 \iint_{R_2} dA = 3 \cdot \text{Area}(R_2)$

The area of $R_1$ is $(3-1)(2-0) = 2 \cdot 2 = 4$.

The area of $R_2$ is $(4-3)(2-0) = 1 \cdot 2 = 2$.

Therefore,

$\iint_R f(x, y) dA = 2 \cdot 4 + 3 \cdot 2 = 8 + 6 = 14$

3. Final Answer

14