Find the value of `int_(0)^(x){t}dt,x inR^(+)`, here {.} denotes fractional part of 'x'.

Answer

To find the value of the integral \( \int_{0}^{x} \{t\} dt \) where \( \{t\} \) denotes the fractional part of \( t \), we can follow these steps:

### Step 1: Understand the Fractional Part
The fractional part of \( t \), denoted as \( \{t\} \), can be expressed as:
\[ \{t\} = t - \lfloor t \rfloor \]
where \( \lfloor t \rfloor \) is the greatest integer less than or equal to \( t \).

### Step 2: Set Up the Integral
We can rewrite the integral as:
\[ \int_{0}^{x} \{t\} dt = \int_{0}^{x} (t - \lfloor t \rfloor) dt \]
This can be separated into two integrals:
\[ \int_{0}^{x} t dt - \int_{0}^{x} \lfloor t \rfloor dt \]

### Step 3: Calculate the First Integral
The first integral is straightforward:
\[ \int_{0}^{x} t dt = \left[ \frac{t^2}{2} \right]_{0}^{x} = \frac{x^2}{2} \]

### Step 4: Analyze the Second Integral
To evaluate \( \int_{0}^{x} \lfloor t \rfloor dt \), we need to consider the behavior of \( \lfloor t \rfloor \) over the interval from \( 0 \) to \( x \).

Assume \( x = n + f \) where \( n = \lfloor x \rfloor \) (the integer part) and \( f = \{x\} \) (the fractional part). The integral can be split into segments:
- From \( 0 \) to \( 1 \): \( \lfloor t \rfloor = 0 \)
- From \( 1 \) to \( 2 \): \( \lfloor t \rfloor = 1 \)
- From \( 2 \) to \( 3 \): \( \lfloor t \rfloor = 2 \)
- ...
- From \( n \) to \( n + f \): \( \lfloor t \rfloor = n \)

Thus, we can express the second integral as:
\[ \int_{0}^{x} \lfloor t \rfloor dt = \int_{0}^{1} 0 dt + \int_{1}^{2} 1 dt + \int_{2}^{3} 2 dt + \ldots + \int_{n}^{n+f} n dt \]

Calculating each segment:
- \( \int_{0}^{1} 0 dt = 0 \)
- \( \int_{1}^{2} 1 dt = 1 \)
- \( \int_{2}^{3} 2 dt = 2 \)
- ...
- \( \int_{n}^{n+f} n dt = n \cdot f \)

The total for the integer segments is:
\[ 0 + 1 + 2 + \ldots + (n-1) + n \cdot f = \frac{(n-1)n}{2} + n \cdot f \]

### Step 5: Combine the Results
Now we can combine the results of both integrals:
\[ \int_{0}^{x} \{t\} dt = \frac{x^2}{2} - \left( \frac{(n-1)n}{2} + n \cdot f \right) \]
This simplifies to:
\[ \int_{0}^{x} \{t\} dt = \frac{x^2}{2} - \frac{(n-1)n}{2} - n \cdot f \]

### Final Result
Thus, the final value of the integral is:
\[ \int_{0}^{x} \{t\} dt = \frac{x^2}{2} - \frac{(n-1)n}{2} - n \cdot \{x\} \]