The area enclosed by the curve
`[x+3y]=[x-2] ` where `x in [3,4]` is :
(where[.] denotes greatest integer function)

Options

A. `2/3`
B. `1/3`
C. `1/4`
D. `1`

Answer

To find the area enclosed by the curve defined by the equation \( [x + 3y] = [x - 2] \) for \( x \in [3, 4] \), we will follow these steps:

### Step 1: Understand the Greatest Integer Function
The greatest integer function, denoted by \([.]\), returns the largest integer less than or equal to the given value. For \(x\) in the interval \([3, 4]\):
- When \(x = 3\), \([x] = 3\)
- When \(x = 4\), \([x] = 4\)
- For \(3 < x < 4\), \([x] = 3\)

### Step 2: Analyze the Equation
The equation can be rewritten as:
\[ [x + 3y] = [x - 2] \]
For \(x\) in \([3, 4]\):
- When \(x = 3\), \([x - 2] = [1] = 1\)
- When \(x = 4\), \([x - 2] = [2] = 2\)

Thus, we need to consider two cases based on the values of \(x\):

1. When \(3 \leq x < 4\):
\[ [x + 3y] = 1 \implies 1 \leq x + 3y < 2 \]
This implies:
\[ 1 - x < 3y < 2 - x \implies \frac{1 - x}{3} < y < \frac{2 - x}{3} \]

2. When \(x = 4\):
\[ [x + 3y] = 2 \implies 2 \leq x + 3y < 3 \]
This implies:
\[ 2 - x < 3y < 3 - x \implies \frac{2 - x}{3} < y < \frac{3 - x}{3} \]

### Step 3: Determine the Limits for \(y\)

For \(x\) in \([3, 4)\):
- When \(x = 3\):
\[ \frac{1 - 3}{3} < y < \frac{2 - 3}{3} \implies -\frac{2}{3} < y < -\frac{1}{3} \]
- When \(x = 4\):
\[ \frac{2 - 4}{3} < y < \frac{3 - 4}{3} \implies -\frac{2}{3} < y < -\frac{1}{3} \]

### Step 4: Calculate the Area
The area can be calculated by integrating the difference of the upper and lower bounds of \(y\) over the interval from \(3\) to \(4\):
\[ \text{Area} = \int_{3}^{4} \left( \frac{2 - x}{3} - \frac{1 - x}{3} \right) \, dx \]
Simplifying the integrand:
\[ \text{Area} = \int_{3}^{4} \left( \frac{2 - x - 1 + x}{3} \right) \, dx = \int_{3}^{4} \frac{1}{3} \, dx \]
Calculating the integral:
\[ \text{Area} = \frac{1}{3} \times (4 - 3) = \frac{1}{3} \]

### Conclusion
The area enclosed by the curve is:
\[ \text{Area} = \frac{1}{3} \text{ square units} \]