If `int_(0)^(x){t}dt=int_(0)^({x})t dt ("where" x gt0 neZ and and {*}` represents fractional part function), then

Options

A. `x in (0,1)`
B. `[x]=1`
C. `x in (1,6)-I`
D. none of these

Explanation

To solve the problem, we need to evaluate the integrals given in the equation:

\[ \int_{0}^{x} \{t\} dt = \int_{0}^{\{x\}} t dt \]

where \(\{t\}\) denotes the fractional part of \(t\) and \(x > 0\) is not an integer.

### Step 1: Understand the fractional part function
The fractional part of a number \(x\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\).

### Step 2: Evaluate the left-hand side integral
We need to evaluate the integral:
\[ \int_{0}^{x} \{t\} dt \]
Since \(x\) is not an integer, we can express \(x\) as:
\[ x = n + f \quad \text{where } n = \lfloor x \rfloor \text{ and } f = \{x\} \]
Thus, we can split the integral from \(0\) to \(x\) into two parts:
\[ \int_{0}^{x} \{t\} dt = \int_{0}^{n} \{t\} dt + \int_{n}^{x} \{t\} dt \]
For \(t\) in the interval \([0, n]\), \(\{t\} = t\), so:
\[ \int_{0}^{n} \{t\} dt = \int_{0}^{n} t dt = \left[\frac{t^2}{2}\right]_{0}^{n} = \frac{n^2}{2} \]
For \(t\) in the interval \([n, x]\), \(\{t\} = t - n\), so:
\[ \int_{n}^{x} \{t\} dt = \int_{n}^{x} (t - n) dt = \int_{n}^{x} t dt - n\int_{n}^{x} dt \]
Calculating these integrals:
\[ \int_{n}^{x} t dt = \left[\frac{t^2}{2}\right]_{n}^{x} = \frac{x^2}{2} - \frac{n^2}{2} \]
\[ \int_{n}^{x} dt = x - n \]
Thus,
\[ \int_{n}^{x} \{t\} dt = \left(\frac{x^2}{2} - \frac{n^2}{2}\right) - n(x - n) = \frac{x^2}{2} - \frac{n^2}{2} - nx + n^2 = \frac{x^2}{2} - nx + \frac{n^2}{2} \]
Combining both parts:
\[ \int_{0}^{x} \{t\} dt = \frac{n^2}{2} + \left(\frac{x^2}{2} - nx + \frac{n^2}{2}\right) = \frac{x^2}{2} - nx + n^2 \]

### Step 3: Evaluate the right-hand side integral
Now we evaluate:
\[ \int_{0}^{\{x\}} t dt \]
Since \(\{x\} = f\):
\[ \int_{0}^{f} t dt = \left[\frac{t^2}{2}\right]_{0}^{f} = \frac{f^2}{2} \]

### Step 4: Set the two integrals equal
Setting the two integrals equal gives us:
\[ \frac{x^2}{2} - nx + n^2 = \frac{f^2}{2} \]
Substituting \(f = x - n\):
\[ \frac{x^2}{2} - nx + n^2 = \frac{(x - n)^2}{2} \]
Expanding the right side:
\[ \frac{x^2}{2} - nx + n^2 = \frac{x^2 - 2nx + n^2}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ x^2 - 2nx + 2n^2 = x^2 - 2nx + n^2 \]
This simplifies to:
\[ 2n^2 = n^2 \implies n^2 = 0 \implies n = 0 \]
Thus, \(x\) must lie in the interval \(0 < x < 1\).

### Conclusion
The solution shows that for \(x > 0\) and \(x\) not being an integer, \(x\) must be in the interval \(0 < x < 1\).