The value of `int_(0)^(2)xd([x]-x)`, is

Options

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

Explanation

To solve the integral \( I = \int_{0}^{2} x \, d([x] - x) \), we will follow these steps:

### Step 1: Understand the expression
The expression \([x]\) is the greatest integer function (also known as the floor function), which gives the largest integer less than or equal to \(x\). The term \([x] - x\) is a piecewise function that changes at integer values of \(x\).

### Step 2: Rewrite the integral using integration by parts
We can use integration by parts, where we let:
- \( u = [x] - x \)
- \( dv = dx \)

Then, we have:
- \( du = d([x]) - dx \)
- \( v = x \)

Using integration by parts, we have:
\[ I = \left[ x([x] - x) \right]_{0}^{2} - \int_{0}^{2} ([x] - x) \, dx \]

### Step 3: Evaluate the boundary term
Now, we evaluate the boundary term:
\[ \left[ x([x] - x) \right]_{0}^{2} = 2([2] - 2) - 0([0] - 0) = 2(2 - 2) - 0 = 0 \]

### Step 4: Evaluate the integral
Next, we need to evaluate the integral \( \int_{0}^{2} ([x] - x) \, dx \). We will split this integral into two parts, from \(0\) to \(1\) and from \(1\) to \(2\):

1. For \(0 \leq x < 1\), \([x] = 0\):
\[ \int_{0}^{1} ([x] - x) \, dx = \int_{0}^{1} (0 - x) \, dx = -\int_{0}^{1} x \, dx = -\left[ \frac{x^2}{2} \right]_{0}^{1} = -\frac{1}{2} \]

2. For \(1 \leq x < 2\), \([x] = 1\):
\[ \int_{1}^{2} ([x] - x) \, dx = \int_{1}^{2} (1 - x) \, dx = \int_{1}^{2} 1 \, dx - \int_{1}^{2} x \, dx \]
The first integral:
\[ \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1 \]
The second integral:
\[ \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{4}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2} \]
Thus,
\[ \int_{1}^{2} ([x] - x) \, dx = 1 - \frac{3}{2} = -\frac{1}{2} \]

### Step 5: Combine the results
Now we combine the results of the two integrals:
\[ \int_{0}^{2} ([x] - x) \, dx = -\frac{1}{2} + (-\frac{1}{2}) = -1 \]

### Step 6: Final result
Putting it all together:
\[ I = 0 - (-1) = 1 \]

Thus, the value of the integral \( I = \int_{0}^{2} x \, d([x] - x) \) is \( \boxed{1} \).