For any real number x, the value of `int_(0)^(x) [x]dx`, is

Options

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

Explanation

To solve the integral \( \int_{0}^{x} \lfloor t \rfloor \, dt \), where \( \lfloor t \rfloor \) is the greatest integer function (also known as the floor function), we will break down the integral into manageable parts based on the value of \( x \).

### Step 1: Understanding the Floor Function
The floor function \( \lfloor t \rfloor \) gives the greatest integer less than or equal to \( t \). For any real number \( x \), we can express \( x \) as:
\[ x = n + f \]
where \( n = \lfloor x \rfloor \) (the integer part) and \( f = x - n \) (the fractional part, where \( 0 \leq f < 1 \)).

### Step 2: Breaking the Integral into Intervals
We will split the integral from \( 0 \) to \( x \) into intervals where \( \lfloor t \rfloor \) is constant:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \int_{0}^{1} \lfloor t \rfloor \, dt + \int_{1}^{2} \lfloor t \rfloor \, dt + \ldots + \int_{n-1}^{n} \lfloor t \rfloor \, dt + \int_{n}^{n+f} \lfloor t \rfloor \, dt \]

### Step 3: Evaluating Each Integral
1. For \( t \) in \( [0, 1) \):
\[ \lfloor t \rfloor = 0 \quad \Rightarrow \quad \int_{0}^{1} \lfloor t \rfloor \, dt = 0 \]

2. For \( t \) in \( [1, 2) \):
\[ \lfloor t \rfloor = 1 \quad \Rightarrow \quad \int_{1}^{2} \lfloor t \rfloor \, dt = \int_{1}^{2} 1 \, dt = 1 \]

3. For \( t \) in \( [2, 3) \):
\[ \lfloor t \rfloor = 2 \quad \Rightarrow \quad \int_{2}^{3} \lfloor t \rfloor \, dt = \int_{2}^{3} 2 \, dt = 2 \]

4. Continuing this pattern, for \( t \) in \( [k, k+1) \) where \( k = 1, 2, \ldots, n-1 \):
\[ \int_{k}^{k+1} \lfloor t \rfloor \, dt = k \]

5. For the last part \( [n, n+f) \):
\[ \lfloor t \rfloor = n \quad \Rightarrow \quad \int_{n}^{n+f} \lfloor t \rfloor \, dt = \int_{n}^{n+f} n \, dt = n \cdot f \]

### Step 4: Summing the Integrals
Now we can sum all these contributions:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = 0 + 1 + 2 + \ldots + (n-1) + n \cdot f \]
The sum \( 1 + 2 + \ldots + (n-1) \) can be calculated using the formula for the sum of the first \( m \) integers:
\[ \sum_{k=1}^{m} k = \frac{m(m+1)}{2} \]
Thus,
\[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2} \]

### Final Result
Putting it all together, we have:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \frac{(n-1)n}{2} + n \cdot f \]
Substituting \( f = x - n \):
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \frac{(n-1)n}{2} + n(x - n) \]
This simplifies to:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \frac{n(n-1)}{2} + nx - n^2 = nx - \frac{n}{2} \]

### Conclusion
Thus, the value of the integral \( \int_{0}^{x} \lfloor t \rfloor \, dt \) is:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \frac{n(n-1)}{2} + n(x - n) \]