`int_(-1)^(1)(x-[x])dx` is not equivalent to

Options

A. `int_(-1)^(1){x}dx`
B. `int_(0)^(2){x}dx`
C. `int_(-1)^(1)[x]dx`
D. 1

Answer

To solve the integral \( \int_{-1}^{1} (x - [x]) \, dx \), where \([x]\) denotes the greatest integer function (or floor function), we can follow these steps:

### Step 1: Understand the expression \( x - [x] \)
The expression \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). Thus, we can rewrite the integral as:
\[ \int_{-1}^{1} (x - [x]) \, dx = \int_{-1}^{1} \{x\} \, dx \]

### Step 2: Analyze the function \( \{x\} \)
The function \( \{x\} \) is defined as:
- For \( x \in [0, 1) \), \( \{x\} = x \)
- For \( x \in [-1, 0) \), \( \{x\} = x + 1 \)

### Step 3: Split the integral
We can split the integral into two parts:
\[ \int_{-1}^{1} \{x\} \, dx = \int_{-1}^{0} \{x\} \, dx + \int_{0}^{1} \{x\} \, dx \]

### Step 4: Evaluate the integral from \(-1\) to \(0\)
For \( x \in [-1, 0) \):
\[ \{x\} = x + 1 \]
Thus,
\[ \int_{-1}^{0} \{x\} \, dx = \int_{-1}^{0} (x + 1) \, dx \]
Calculating this gives:
\[ = \int_{-1}^{0} x \, dx + \int_{-1}^{0} 1 \, dx \]
Calculating each part:
- \( \int_{-1}^{0} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{0} = 0 - \frac{(-1)^2}{2} = -\frac{1}{2} \)
- \( \int_{-1}^{0} 1 \, dx = [x]_{-1}^{0} = 0 - (-1) = 1 \)

Thus,
\[ \int_{-1}^{0} (x + 1) \, dx = -\frac{1}{2} + 1 = \frac{1}{2} \]

### Step 5: Evaluate the integral from \(0\) to \(1\)
For \( x \in [0, 1) \):
\[ \{x\} = x \]
Thus,
\[ \int_{0}^{1} \{x\} \, dx = \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - 0 = \frac{1}{2} \]

### Step 6: Combine the results
Now, we combine the results from both intervals:
\[ \int_{-1}^{1} \{x\} \, dx = \int_{-1}^{0} \{x\} \, dx + \int_{0}^{1} \{x\} \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

### Conclusion
Thus, the value of the integral \( \int_{-1}^{1} (x - [x]) \, dx \) is \( 1 \).