What is
`int_(-2)^(2) xdx -int_(-2)^(2) [x]dx`
equal to , where [.] is the greatest integer function ?

Options

A. `0`
B. `1`
C. `2`
D. `4`

Explanation

To solve the problem, we need to evaluate the expression:

\[ \int_{-2}^{2} x \, dx - \int_{-2}^{2} [x] \, dx \]

where \([x]\) is the greatest integer function (also known as the floor function).

### Step 1: Evaluate \(\int_{-2}^{2} x \, dx\)

The integral of \(x\) can be calculated as follows:

\[ \int x \, dx = \frac{x^2}{2} \]

Now, we will evaluate this from \(-2\) to \(2\):

\[ \int_{-2}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{-2}^{2} = \frac{2^2}{2} - \frac{(-2)^2}{2} = \frac{4}{2} - \frac{4}{2} = 2 - 2 = 0 \]

### Step 2: Evaluate \(\int_{-2}^{2} [x] \, dx\)

Next, we need to evaluate the integral of the greatest integer function \([x]\) from \(-2\) to \(2\). The greatest integer function \([x]\) takes on different constant values over the intervals:

- For \(x \in [-2, -1)\), \([x] = -2\)
- For \(x \in [-1, 0)\), \([x] = -1\)
- For \(x \in [0, 1)\), \([x] = 0\)
- For \(x \in [1, 2)\), \([x] = 1\)
- At \(x = 2\), \([x] = 2\)

Thus, we can break the integral into segments:

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

Calculating each integral:

1. \(\int_{-2}^{-1} (-2) \, dx = -2 \cdot (-1 - (-2)) = -2 \cdot 1 = -2\)
2. \(\int_{-1}^{0} (-1) \, dx = -1 \cdot (0 - (-1)) = -1 \cdot 1 = -1\)
3. \(\int_{0}^{1} (0) \, dx = 0\)
4. \(\int_{1}^{2} (1) \, dx = 1 \cdot (2 - 1) = 1\)

Now, summing these results:

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

### Step 3: Combine the results

Now we can substitute back into our original expression:

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

### Final Answer

Thus, the value of the expression is:

\[ \boxed{2} \]