If `n in N`, the value of `int_(0)^(n) [x] dx` (where [x] is the greatest integer function) is

Options

A. `(n(n+1))/(2)`
B. `(n(n-1))/(2)`
C. `n(n-1)`
D. none of these

Answer

To solve the integral \( \int_{0}^{n} [x] \, dx \), where \([x]\) is the greatest integer function (also known as the floor function), we can break the integral into segments based on the behavior of the greatest integer function.

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
The function \([x]\) takes the value of the greatest integer less than or equal to \(x\). Therefore, within the interval \([0, n)\), the function \([x]\) will take the values:
- \([x] = 0\) for \(0 \leq x < 1\)
- \([x] = 1\) for \(1 \leq x < 2\)
- \([x] = 2\) for \(2 \leq x < 3\)
- ...
- \([x] = n-1\) for \(n-1 \leq x < n\)

2. **Breaking the Integral**:
We can break the integral from \(0\) to \(n\) into separate integrals for each interval:
\[ \int_{0}^{n} [x] \, dx = \int_{0}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \ldots + \int_{n-1}^{n} (n-1) \, dx \]

3. **Calculating Each Integral**:
- The first integral:
\[ \int_{0}^{1} 0 \, dx = 0 \]
- The second integral:
\[ \int_{1}^{2} 1 \, dx = 1 \cdot (2 - 1) = 1 \]
- The third integral:
\[ \int_{2}^{3} 2 \, dx = 2 \cdot (3 - 2) = 2 \]
- Continuing this pattern, we find:
\[ \int_{k}^{k+1} k \, dx = k \cdot (k+1 - k) = k \quad \text{for } k = 1, 2, \ldots, n-1 \]

4. **Summing the Integrals**:
The total integral can now be expressed as:
\[ \int_{0}^{n} [x] \, dx = 0 + 1 + 2 + \ldots + (n-1) \]
This is the sum of the first \(n-1\) natural numbers.

5. **Using the Formula for the Sum of Natural Numbers**:
The sum of the first \(m\) natural numbers is given by:
\[ S = \frac{m(m + 1)}{2} \]
For \(m = n-1\), we have:
\[ S = \frac{(n-1)n}{2} \]

6. **Final Result**:
Therefore, the value of the integral is:
\[ \int_{0}^{n} [x] \, dx = \frac{(n-1)n}{2} \]