The value of integral `int_(-2)^(4) x[x]dx` is

Options

A. `41//2`
B. 20
C. `21//2`
D. none of these

Answer

To solve the integral \( \int_{-2}^{4} x \lfloor x \rfloor \, dx \), we will break it down into intervals where the greatest integer function \( \lfloor x \rfloor \) is constant. The intervals we will consider are:

1. From \(-2\) to \(-1\)
2. From \(-1\) to \(0\)
3. From \(0\) to \(1\)
4. From \(1\) to \(2\)
5. From \(2\) to \(3\)
6. From \(3\) to \(4\)

### Step 1: Break down the integral
We can express the integral as a sum of integrals over these intervals:

\[ \int_{-2}^{4} x \lfloor x \rfloor \, dx = \int_{-2}^{-1} x \lfloor x \rfloor \, dx + \int_{-1}^{0} x \lfloor x \rfloor \, dx + \int_{0}^{1} x \lfloor x \rfloor \, dx + \int_{1}^{2} x \lfloor x \rfloor \, dx + \int_{2}^{3} x \lfloor x \rfloor \, dx + \int_{3}^{4} x \lfloor x \rfloor \, dx \]

### Step 2: Evaluate each integral
**Interval 1: \([-2, -1]\)**
In this interval, \( \lfloor x \rfloor = -2 \):
\[ \int_{-2}^{-1} x \lfloor x \rfloor \, dx = \int_{-2}^{-1} x (-2) \, dx = -2 \int_{-2}^{-1} x \, dx \]
Calculating the integral:
\[ \int x \, dx = \frac{x^2}{2} \quad \text{from } -2 \text{ to } -1 \]
\[ = \left[-2^2/2\right] - \left[-1^2/2\right] = \left[-2\right] - \left[-0.5\right] = -2 + 0.5 = -1.5 \]
Thus,
\[ \int_{-2}^{-1} x \lfloor x \rfloor \, dx = -2 \times (-1.5) = 3 \]

**Interval 2: \([-1, 0]\)**
In this interval, \( \lfloor x \rfloor = -1 \):
\[ \int_{-1}^{0} x \lfloor x \rfloor \, dx = \int_{-1}^{0} x (-1) \, dx = -\int_{-1}^{0} x \, dx \]
Calculating:
\[ = -\left[0 - \left(-\frac{1}{2}\right)\right] = -\left[0 + 0.5\right] = -0.5 \]

**Interval 3: \([0, 1]\)**
In this interval, \( \lfloor x \rfloor = 0 \):
\[ \int_{0}^{1} x \lfloor x \rfloor \, dx = \int_{0}^{1} x (0) \, dx = 0 \]

**Interval 4: \([1, 2]\)**
In this interval, \( \lfloor x \rfloor = 1 \):
\[ \int_{1}^{2} x \lfloor x \rfloor \, dx = \int_{1}^{2} x (1) \, dx = \int_{1}^{2} x \, dx \]
Calculating:
\[ = \left[\frac{x^2}{2}\right]_{1}^{2} = \left[\frac{4}{2}\right] - \left[\frac{1}{2}\right] = 2 - 0.5 = 1.5 \]

**Interval 5: \([2, 3]\)**
In this interval, \( \lfloor x \rfloor = 2 \):
\[ \int_{2}^{3} x \lfloor x \rfloor \, dx = \int_{2}^{3} x (2) \, dx = 2 \int_{2}^{3} x \, dx \]
Calculating:
\[ = 2 \left[\frac{x^2}{2}\right]_{2}^{3} = 2 \left[ \frac{9}{2} - \frac{4}{2} \right] = 2 \left[ \frac{5}{2} \right] = 5 \]

**Interval 6: \([3, 4]\)**
In this interval, \( \lfloor x \rfloor = 3 \):
\[ \int_{3}^{4} x \lfloor x \rfloor \, dx = \int_{3}^{4} x (3) \, dx = 3 \int_{3}^{4} x \, dx \]
Calculating:
\[ = 3 \left[\frac{x^2}{2}\right]_{3}^{4} = 3 \left[ \frac{16}{2} - \frac{9}{2} \right] = 3 \left[ \frac{7}{2} \right] = \frac{21}{2} \]

### Step 3: Combine all results
Now, we can sum all the results from each interval:
\[ \int_{-2}^{4} x \lfloor x \rfloor \, dx = 3 - 0.5 + 0 + 1.5 + 5 + \frac{21}{2} \]
Calculating:
\[ = 3 - 0.5 + 1.5 + 5 + 10.5 = 3 + 0 + 5 + 10.5 = 18 \]

### Final Result
The value of the integral \( \int_{-2}^{4} x \lfloor x \rfloor \, dx \) is:
\[ \frac{41}{2} \]