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 given by different piecewise functions in problems 1, 2, and 3.

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 given by different piecewise functions in problems 1, 2, and

3.

2. Solution Steps

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

We can split the region $R$ into two subregions: $R_1 = \{(x, y): 1 \le x < 3, 0 \le y \le 2\}$ and $R_2 = \{(x, y): 3 \le x \le 4, 0 \le y \le 2\}$.

Then

$$\iint_R f(x, y) dA = \iint_{R_1} 2 dA + \iint_{R_2} 3 dA$$

$$\iint_{R_1} 2 dA = 2 \int_1^3 \int_0^2 dy dx = 2 \int_1^3 [y]_0^2 dx = 2 \int_1^3 2 dx = 4 \int_1^3 dx = 4[x]_1^3 = 4(3-1) = 4(2) = 8$$

$$\iint_{R_2} 3 dA = 3 \int_3^4 \int_0^2 dy dx = 3 \int_3^4 [y]_0^2 dx = 3 \int_3^4 2 dx = 6 \int_3^4 dx = 6[x]_3^4 = 6(4-3) = 6(1) = 6$$

$$\iint_R f(x, y) dA = 8 + 6 = 14$$

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

We can split the region $R$ into two subregions: $R_1 = \{(x, y): 1 \le x \le 4, 0 \le y < 1\}$ and $R_2 = \{(x, y): 1 \le x \le 4, 1 \le y \le 2\}$.

Then

$$\iint_R f(x, y) dA = \iint_{R_1} (-1) dA + \iint_{R_2} 2 dA$$

$$\iint_{R_1} (-1) dA = -1 \int_1^4 \int_0^1 dy dx = -1 \int_1^4 [y]_0^1 dx = -1 \int_1^4 1 dx = -\int_1^4 dx = -[x]_1^4 = -(4-1) = -3$$

$$\iint_{R_2} 2 dA = 2 \int_1^4 \int_1^2 dy dx = 2 \int_1^4 [y]_1^2 dx = 2 \int_1^4 1 dx = 2[x]_1^4 = 2(4-1) = 2(3) = 6$$

$$\iint_R f(x, y) dA = -3 + 6 = 3$$

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

We can split the region $R$ into three subregions: $R_1 = \{(x, y): 1 \le x < 3, 0 \le y < 1\}$, $R_2 = \{(x, y): 1 \le x < 3, 1 \le y \le 2\}$, and $R_3 = \{(x, y): 3 \le x \le 4, 0 \le y \le 2\}$.

Then

$$\iint_R f(x, y) dA = \iint_{R_1} 2 dA + \iint_{R_2} 1 dA + \iint_{R_3} 3 dA$$

$$\iint_{R_1} 2 dA = 2 \int_1^3 \int_0^1 dy dx = 2 \int_1^3 [y]_0^1 dx = 2 \int_1^3 1 dx = 2[x]_1^3 = 2(3-1) = 2(2) = 4$$

$$\iint_{R_2} 1 dA = \int_1^3 \int_1^2 dy dx = \int_1^3 [y]_1^2 dx = \int_1^3 1 dx = [x]_1^3 = 3-1 = 2$$

$$\iint_{R_3} 3 dA = 3 \int_3^4 \int_0^2 dy dx = 3 \int_3^4 [y]_0^2 dx = 3 \int_3^4 2 dx = 6 \int_3^4 dx = 6[x]_3^4 = 6(4-3) = 6(1) = 6$$

$$\iint_R f(x, y) dA = 4 + 2 + 6 = 12$$

3. Final Answer

1. 14

2. 3

3. 12