If `int_0^x[x]dx=int_0^([x]) xdx,x !in` integer (where, [*] and {*} denotes the greatest integer and fractional parts respectively,then the value of `4{x}` is equal to ...

Answer

To solve the given problem, we need to evaluate the integral equation and find the value of \(4\{x\}\), where \(\{x\}\) is the fractional part of \(x\) and \([x]\) is the greatest integer less than or equal to \(x\).

### Step-by-Step Solution:

1. **Understanding the Integral**:
We start with the equation:
\[ \int_0^x [x] \, dx = \int_0^{[x]} x \, dx \]
where \(x\) is not an integer.

2. **Evaluating the Left Side**:
The left-hand side involves the greatest integer function \([x]\). We can express \(x\) as \(n + f\), where \(n = [x]\) (the integer part) and \(f = \{x\}\) (the fractional part). Thus, we can break the integral into segments:
\[ \int_0^x [x] \, dx = \int_0^1 0 \, dx + \int_1^2 1 \, dx + \int_2^3 2 \, dx + \ldots + \int_{n-1}^n (n-1) \, dx + \int_n^{n+f} n \, dx \]
This simplifies to:
\[ = 0 + 1 + 2 + \ldots + (n-1) + n \cdot f \]
The sum of the first \(n-1\) integers is:
\[ \frac{(n-1)n}{2} \]
Therefore, the left-hand side becomes:
\[ \frac{(n-1)n}{2} + nf \]

3. **Evaluating the Right Side**:
The right-hand side is:
\[ \int_0^{[x]} x \, dx = \int_0^n x \, dx = \left[\frac{x^2}{2}\right]_0^n = \frac{n^2}{2} \]

4. **Setting the Two Sides Equal**:
We equate the two sides:
\[ \frac{(n-1)n}{2} + nf = \frac{n^2}{2} \]

5. **Simplifying the Equation**:
Rearranging gives:
\[ nf = \frac{n^2}{2} - \frac{(n-1)n}{2} \]
Simplifying the right-hand side:
\[ nf = \frac{n^2 - (n^2 - n)}{2} = \frac{n}{2} \]
Thus, we have:
\[ f = \frac{1}{2} \]

6. **Finding \(4\{x\}\)**:
Since \(f = \{x\}\), we find:
\[ 4\{x\} = 4f = 4 \cdot \frac{1}{2} = 2 \]

### Final Answer:
The value of \(4\{x\}\) is \(2\).