Find the area enclosed by the curve `[x]+[y]-4` in 1st quadrant (where [.] denotes greatest integer function).

Answer

To find the area enclosed by the curve \([x] + [y] - 4 = 0\) in the first quadrant, we will analyze the equation step by step.

### Step 1: Understanding the Equation
The equation \([x] + [y] - 4 = 0\) can be rewritten as \([x] + [y] = 4\). Here, \([x]\) and \([y]\) represent the greatest integer less than or equal to \(x\) and \(y\), respectively.

**Hint:** The greatest integer function, \([x]\), takes integer values, so we can consider different cases based on the possible integer values of \([x]\) and \([y]\).

### Step 2: Case Analysis
We will analyze the cases for \([x]\) and \([y]\) that satisfy the equation \([x] + [y] = 4\).

- **Case 1:** \([x] = 0\)
Then \([y] = 4\).
This means \(0 \leq x < 1\) and \(4 \leq y < 5\).
This gives us a rectangle in the first quadrant with vertices at \((0, 4)\), \((1, 4)\), \((1, 5)\), and \((0, 5)\).

- **Case 2:** \([x] = 1\)
Then \([y] = 3\).
This means \(1 \leq x < 2\) and \(3 \leq y < 4\).
This gives us another rectangle with vertices at \((1, 3)\), \((2, 3)\), \((2, 4)\), and \((1, 4)\).

- **Case 3:** \([x] = 2\)
Then \([y] = 2\).
This means \(2 \leq x < 3\) and \(2 \leq y < 3\).
This gives us another rectangle with vertices at \((2, 2)\), \((3, 2)\), \((3, 3)\), and \((2, 3)\).

- **Case 4:** \([x] = 3\)
Then \([y] = 1\).
This means \(3 \leq x < 4\) and \(1 \leq y < 2\).
This gives us another rectangle with vertices at \((3, 1)\), \((4, 1)\), \((4, 2)\), and \((3, 2)\).

- **Case 5:** \([x] = 4\)
Then \([y] = 0\).
This means \(4 \leq x < 5\) and \(0 \leq y < 1\).
This gives us another rectangle with vertices at \((4, 0)\), \((5, 0)\), \((5, 1)\), and \((4, 1)\).

### Step 3: Calculating the Area
Now, we will calculate the area of each rectangle.

1. **Area of Case 1:**
Width = 1, Height = 1 (from 4 to 5)
Area = \(1 \times 1 = 1\)

2. **Area of Case 2:**
Width = 1, Height = 1 (from 3 to 4)
Area = \(1 \times 1 = 1\)

3. **Area of Case 3:**
Width = 1, Height = 1 (from 2 to 3)
Area = \(1 \times 1 = 1\)

4. **Area of Case 4:**
Width = 1, Height = 1 (from 1 to 2)
Area = \(1 \times 1 = 1\)

5. **Area of Case 5:**
Width = 1, Height = 1 (from 0 to 1)
Area = \(1 \times 1 = 1\)

### Step 4: Total Area
Now, we sum the areas of all the rectangles:
\[ \text{Total Area} = 1 + 1 + 1 + 1 + 1 = 5 \text{ square units} \]

### Final Answer
The area enclosed by the curve \([x] + [y] - 4 = 0\) in the first quadrant is \(5\) square units.

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