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

### Step 1: Express the fractional part
The fractional part of \( t \) 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: Rewrite the integral
Using the expression for the fractional part, we can rewrite the integral:
\[ I = \int_{0}^{x} \{t\} \, dt = \int_{0}^{x} (t - \lfloor t \rfloor) \, dt \]
This can be split into two separate integrals:
\[ I = \int_{0}^{x} t \, dt - \int_{0}^{x} \lfloor t \rfloor \, dt \]

### 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: Calculate the second integral
To evaluate the second integral \( \int_{0}^{x} \lfloor t \rfloor \, dt \), we need to consider the behavior of \( \lfloor t \rfloor \) over the interval \( [0, x] \).

Let \( n = \lfloor x \rfloor \) (the greatest integer less than or equal to \( x \)). The integral can be computed as a sum of areas of rectangles:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \sum_{k=0}^{n-1} \int_{k}^{k+1} k \, dt + \int_{n}^{x} n \, dt \]
Calculating these integrals:
1. For \( k = 0 \) to \( n-1 \):
\[ \int_{k}^{k+1} k \, dt = k \cdot (1) = k \]
Thus, the sum becomes:
\[ \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \]
2. For the last part from \( n \) to \( x \):
\[ \int_{n}^{x} n \, dt = n(x - n) \]

Putting it all together:
\[ \int_{0}^{x} \lfloor t \rfloor \, dt = \frac{(n-1)n}{2} + n(x - n) = \frac{n^2 - n + 2nx - 2n^2}{2} = \frac{2nx - n^2 - n}{2} \]

### Step 5: Combine results
Now we can combine the results from Step 3 and Step 4:
\[ I = \frac{x^2}{2} - \left( \frac{n^2 - n + 2nx - n^2}{2} \right) = \frac{x^2}{2} - \frac{2nx - n}{2} = \frac{x^2 - 2nx + n}{2} \]

### Final Expression
Thus, the final expression for the integral is:
\[ I = \frac{x^2 - 2n x + n}{2} \]
This can also be simplified to:
\[ I = \frac{(x - n)^2}{2} \]
where \( n = \lfloor x \rfloor \).