`int_0^(4) [x[x+[x+[x]]]]` dx is equal to where `[.]` is greatest integer functing

Options

A. `4`
B. `12`
C. `24`
D. None of these

Explanation

To solve the integral \( \int_0^4 [x[x+[x+[x]]]] \, dx \), where \([.]\) denotes the greatest integer function, we will break the integral into segments where the greatest integer function remains constant.

### Step-by-Step Solution:

1. **Identify the intervals**: The greatest integer function \([x]\) changes at every integer. Therefore, we will break the integral into the intervals \([0, 1)\), \([1, 2)\), \([2, 3)\), and \([3, 4)\).

2. **Evaluate the integral on each interval**:
- **Interval [0, 1)**:
\[ [x] = 0 \quad \text{for } x \in [0, 1) \]
Thus, the expression simplifies to:
\[ [x[x+[x+[x]]]] = [x[0 + 0 + 0]] = [0] = 0 \]
Therefore,
\[ \int_0^1 [x[x+[x+[x]]]] \, dx = \int_0^1 0 \, dx = 0 \]

- **Interval [1, 2)**:
\[ [x] = 1 \quad \text{for } x \in [1, 2) \]
Thus, the expression simplifies to:
\[ [x[x+[x+[x]]]] = [x[1 + 1 + 1]] = [3x] \]
Since \(3x\) ranges from \(3\) to \(6\) in this interval, we have:
\[ [3x] = 3 \quad \text{for } x \in [1, \frac{2}{3}) \quad \text{and} \quad [3x] = 4 \quad \text{for } x \in [\frac{2}{3}, 2) \]
Therefore,
\[ \int_1^2 [3x] \, dx = \int_1^{\frac{2}{3}} 3 \, dx + \int_{\frac{2}{3}}^2 4 \, dx \]
Evaluating these integrals gives:
\[ \int_1^{\frac{2}{3}} 3 \, dx = 3 \left(\frac{2}{3} - 1\right) = -\frac{3}{3} = -1 \]
\[ \int_{\frac{2}{3}}^2 4 \, dx = 4 \left(2 - \frac{2}{3}\right) = 4 \cdot \frac{4}{3} = \frac{16}{3} \]
Thus,
\[ \int_1^2 [3x] \, dx = -1 + \frac{16}{3} = \frac{13}{3} \]

- **Interval [2, 3)**:
\[ [x] = 2 \quad \text{for } x \in [2, 3) \]
Thus, the expression simplifies to:
\[ [x[x+[x+[x]]]] = [x[2 + 2 + 2]] = [6x] \]
Therefore,
\[ \int_2^3 [6x] \, dx = \int_2^3 12 \, dx = 12(3 - 2) = 12 \]

- **Interval [3, 4)**:
\[ [x] = 3 \quad \text{for } x \in [3, 4) \]
Thus, the expression simplifies to:
\[ [x[x+[x+[x]]]] = [x[3 + 3 + 3]] = [9x] \]
Therefore,
\[ \int_3^4 [9x] \, dx = \int_3^4 9 \, dx = 9(4 - 3) = 9 \]

3. **Combine the results**:
\[ \int_0^4 [x[x+[x+[x]]]] \, dx = 0 + \frac{13}{3} + 12 + 9 \]
Converting \(12\) and \(9\) to fractions:
\[ 12 = \frac{36}{3}, \quad 9 = \frac{27}{3} \]
Thus,
\[ \int_0^4 [x[x+[x+[x]]]] \, dx = 0 + \frac{13}{3} + \frac{36}{3} + \frac{27}{3} = \frac{76}{3} \]

### Final Answer:
\[ \int_0^4 [x[x+[x+[x]]]] \, dx = \frac{76}{3} \]