The value of l =`int_0^3([x]+[x+1/3]+[x+2/3])dx` where [.] denotes the greatest integer function is equal to

Options

A. 10
B. 11
C. 12
D. none of these

Answer

To solve the integral \( l = \int_0^3 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx \), where \([.]\) denotes the greatest integer function, we will break this integral into parts based on the intervals defined by the greatest integer function.

### Step 1: Break the Integral into Intervals
The function \([x]\) changes its value at integer points. Therefore, we will evaluate the integral from \(0\) to \(3\) by breaking it into intervals:
- From \(0\) to \(1\)
- From \(1\) to \(2\)
- From \(2\) to \(3\)

Thus, we can write:
\[ l = \int_0^1 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx + \int_1^2 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx + \int_2^3 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx \]

### Step 2: Evaluate Each Integral
#### Interval \(0 \leq x < 1\)
In this interval:
- \([x] = 0\)
- \([x + \frac{1}{3}] = 0\)
- \([x + \frac{2}{3}] = 0\)

Thus,
\[ \int_0^1 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx = \int_0^1 (0 + 0 + 0) dx = 0 \]

#### Interval \(1 \leq x < 2\)
In this interval:
- \([x] = 1\)
- \([x + \frac{1}{3}] = 1\)
- \([x + \frac{2}{3}] = 1\)

Thus,
\[ \int_1^2 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx = \int_1^2 (1 + 1 + 1) dx = \int_1^2 3 dx = 3 \cdot (2 - 1) = 3 \]

#### Interval \(2 \leq x < 3\)
In this interval:
- \([x] = 2\)
- \([x + \frac{1}{3}] = 2\)
- \([x + \frac{2}{3}] = 2\)

Thus,
\[ \int_2^3 \left( [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}] \right) dx = \int_2^3 (2 + 2 + 2) dx = \int_2^3 6 dx = 6 \cdot (3 - 2) = 6 \]

### Step 3: Combine the Results
Now we combine the results from all intervals:
\[ l = 0 + 3 + 6 = 9 \]

### Final Answer
Thus, the value of \( l \) is \( 9 \).