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 solve the problem, we need to find the area enclosed by the curve defined by the equation \([x + 3y] = [x - 2]\) for \(x\) in the interval \([3, 4]\). The notation \([.]\) denotes the greatest integer function (or floor function).

### Step 1: Analyze the equation

The equation \([x + 3y] = [x - 2]\) implies that the greatest integer value of \(x + 3y\) is equal to the greatest integer value of \(x - 2\).

### Step 2: Determine the values of \([x - 2]\)

For \(x\) in the interval \([3, 4]\):
- When \(x = 3\), \([x - 2] = [3 - 2] = [1] = 1\).
- When \(x = 4\), \([x - 2] = [4 - 2] = [2] = 2\).

Thus, as \(x\) varies from \(3\) to \(4\), \([x - 2]\) takes the values \(1\) and \(2\).

### Step 3: Break down the intervals

1. For \(3 \leq x < 4\):
- \([x - 2] = 1\)
- Therefore, \([x + 3y] = 1\).

2. At \(x = 4\):
- \([x - 2] = 2\)
- Therefore, \([x + 3y] = 2\).

### Step 4: Solve for \(y\)

#### Case 1: When \(3 \leq x < 4\)

From \([x + 3y] = 1\):
\[ 1 \leq x + 3y < 2 \]
This can be rearranged to find \(y\):
\[ 1 - x \leq 3y < 2 - x \]
Dividing by \(3\):
\[ \frac{1 - x}{3} \leq y < \frac{2 - x}{3} \]

#### Case 2: When \(x = 4\)

From \([x + 3y] = 2\):
\[ 2 \leq x + 3y < 3 \]
This can be rearranged to find \(y\):
\[ 2 - x \leq 3y < 3 - x \]
Substituting \(x = 4\):
\[ 2 - 4 \leq 3y < 3 - 4 \]
\[ -2 \leq 3y < -1 \]
Dividing by \(3\):
\[ -\frac{2}{3} \leq y < -\frac{1}{3} \]

### Step 5: Determine the area

Now we need to find the area between the two curves defined by \(y\) in the intervals we derived.

1. For \(3 \leq x < 4\):
- The lower bound for \(y\) is \(\frac{1 - x}{3}\).
- The upper bound for \(y\) is \(\frac{2 - x}{3}\).

The area \(A\) can be calculated using the integral:
\[ A = \int_{3}^{4} \left( \frac{2 - x}{3} - \frac{1 - x}{3} \right) dx \]
\[ = \int_{3}^{4} \left( \frac{2 - x - 1 + x}{3} \right) dx \]
\[ = \int_{3}^{4} \frac{1}{3} dx \]
\[ = \frac{1}{3} \left[ x \right]_{3}^{4} \]
\[ = \frac{1}{3} (4 - 3) = \frac{1}{3} \]

### Final Answer

The area enclosed by the curve is \(\frac{1}{3}\).