The value of `int({[x]})dx` where {.} and [.] denotes the fractional part of x and greatest integer function equals

Options

A. 0
B. 1
C. 2
D. -1

Answer

To solve the integral \( \int \{x\} [x] \, dx \), where \( \{x\} \) denotes the fractional part of \( x \) and \( [x] \) denotes the greatest integer function (or floor function), we can follow these steps:

### Step 1: Understand the Functions
- The greatest integer function \( [x] \) gives the largest integer less than or equal to \( x \).
- The fractional part function \( \{x\} \) is defined as \( \{x\} = x - [x] \).

### Step 2: Analyze the Integral
We can express the integral as:
\[ \int \{x\} [x] \, dx = \int (x - [x]) [x] \, dx \]
However, since \( [x] \) is constant over the intervals \( [n, n+1) \) for \( n \in \mathbb{Z} \), we can break the integral into intervals.

### Step 3: Set Up the Integral Over Intervals
For \( n \leq x < n+1 \):
- \( [x] = n \)
- \( \{x\} = x - n \)

Thus, we can write:
\[ \int_n^{n+1} \{x\} [x] \, dx = \int_n^{n+1} (x - n) n \, dx = n \int_n^{n+1} (x - n) \, dx \]

### Step 4: Evaluate the Integral
Now, we compute:
\[ \int_n^{n+1} (x - n) \, dx \]
Let \( u = x - n \), then \( du = dx \) and the limits change from \( 0 \) to \( 1 \):
\[ \int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2} \]

Thus,
\[ \int_n^{n+1} (x - n) \, dx = \frac{1}{2} \]

### Step 5: Combine the Results
Now substituting back:
\[ \int_n^{n+1} \{x\} [x] \, dx = n \cdot \frac{1}{2} = \frac{n}{2} \]

### Step 6: Sum Over All Intervals
To find the total integral over all intervals, we sum from \( n = 0 \) to \( N \) (where \( N \) is the upper limit of integration):
\[ \int_0^N \{x\} [x] \, dx = \sum_{n=0}^{N-1} \frac{n}{2} = \frac{1}{2} \sum_{n=0}^{N-1} n = \frac{1}{2} \cdot \frac{(N-1)N}{2} = \frac{(N-1)N}{4} \]

### Final Result
Thus, the value of the integral \( \int \{x\} [x] \, dx \) is:
\[ \frac{(N-1)N}{4} + C \]
where \( C \) is the constant of integration.