1. Diketahui R = {(x,y): 15x54, 0 <= y <= 2 \ . Hitunglah  integral lipat dua R f(x, y)dA, dengan f adalah f(x) 2, 1 <= x <= 4, 0 <= y < 1 f(x,y)=3, 1 <= x < 3, 1 <= y <= 2 1. 3 <= x < 4, 1 <= y <= 2

To solve this problem, we are given a region \( R \) and a piecewise-defined function \( f(x, y) \), and we need to compute the double integral of \( f(x, y) \) over this region.

The function \( f(x, y) \) is given as:
\[
f(x, y) = 
\begin{cases} 
2 & \text{if } 1 \leq x \leq 4 \text{ and } 0 \leq y < 1 \\
3 & \text{if } 1 \leq x < 3 \text{ and } 1 \leq y \leq 2 \\
1 & \text{if } 3 \leq x < 4 \text{ and } 1 \leq y \leq 2
\end{cases}
\]

### Step 1: Set up the double integral
The region \( R \) is defined by:
\[
R = \{(x, y): 1 \leq x \leq 4, 0 \leq y \leq 2\}
\]
Thus, the limits for \( x \) and \( y \) are:
- \( x \in [1, 4] \)
- \( y \in [0, 2] \)

Since the function \( f(x, y) \) is piecewise, we will compute the integral over different regions of \( R \).

### Step 2: Break the region and compute integrals

#### Region 1: \( 1 \leq x \leq 4 \) and \( 0 \leq y < 1 \)
Here, \( f(x, y) = 2 \). The integral over this region is:
\[
\int_1^4 \int_0^1 2 \, dy \, dx
\]
The inner integral is:
\[
\int_0^1 2 \, dy = 2 \cdot 1 = 2
\]
Now integrate with respect to \( x \):
\[
\int_1^4 2 \, dx = 2(x) \big|_1^4 = 2(4 - 1) = 6
\]

#### Region 2: \( 1 \leq x < 3 \) and \( 1 \leq y \leq 2 \)
Here, \( f(x, y) = 3 \). The integral over this region is:
\[
\int_1^3 \int_1^2 3 \, dy \, dx
\]
The inner integral is:
\[
\int_1^2 3 \, dy = 3 \cdot 1 = 3
\]
Now integrate with respect to \( x \):
\[
\int_1^3 3 \, dx = 3(x) \big|_1^3 = 3(3 - 1) = 6
\]

#### Region 3: \( 3 \leq x < 4 \) and \( 1 \leq y \leq 2 \)
Here, \( f(x, y) = 1 \). The integral over this region is:
\[
\int_3^4 \int_1^2 1 \, dy \, dx
\]
The inner integral is:
\[
\int_1^2 1 \, dy = 1 \cdot 1 = 1
\]
Now integrate with respect to \( x \):
\[
\int_3^4 1 \, dx = 1(x) \big|_3^4 = 1(4 - 3) = 1
\]

### Step 3: Sum the results
Now sum the results of the integrals over each region:
\[
6 + 6 + 1 = 13
\]

### Final Answer:
The value of the double integral is \( \boxed{13} \).

Do you have any questions, or would you like further clarification on any step?

---

Here are 5 related questions to expand your understanding:
1. How would the integral change if the boundaries for \( y \) were different?
2. Can you find the integral if the function \( f(x, y) \) had different values in each region?
3. What happens if \( f(x, y) \) is not piecewise but continuous?
4. How would you approach this problem using polar coordinates if the region was circular?
5. How do you interpret the result of the double integral geometrically?

**Tip**: For piecewise functions, break the integration into separate regions where the function has different expressions.

Diketahui R = {(x,y): 15x54, 0 <= y <= 2}. Hitunglah integral lipat dua R f(x, y)dA, dengan f adalah f(x) 2, 1 <= x <= 4, 0 <= y < 1, f(x,y)=3, 1 <= x < 3, 1 <= y <= 2, dan f(x,y)=1, 3 <= x < 4, 1 <= y <= 2.

Learn how to compute a double integral of a piecewise-defined function f(x, y) over the region R = {(x,y): 1 ≤ x ≤ 4, 0 ≤ y ≤ 2}. This solution walks through the step-by-step process, breaking the region into parts and calculating integrals for each. Suitable for students in undergraduate calculus.

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