Math Problem Statement

  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

Solution

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

The function f(x,y)f(x, y) is given as:

\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.

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Math Problem Analysis

Mathematical Concepts

Double Integration
Piecewise Functions
Multivariable Calculus

Formulas

Double integral formula: ∬_R f(x, y) dA
Integral of constant functions
Piecewise integration over different regions

Theorems

Fundamental Theorem of Calculus (Multivariable)
Properties of Definite Integrals

Suitable Grade Level

Undergraduate Calculus