if `I = int_(-3) ^2 (|x+1|+|x+2|+|x-1|)dx`

Options

A. `(47)/(2)`
B. `(45)/(2)`
C. `(37)/(2)`
D. `(39)/(2)`

Answer

To solve the integral \( I = \int_{-3}^{2} (|x+1| + |x+2| + |x-1|) \, dx \), we first need to analyze the absolute value expressions to determine the points where they change.

### Step 1: Identify the critical points
The expressions \( |x+1| \), \( |x+2| \), and \( |x-1| \) change at the points:
- \( x = -2 \) (for \( |x+2| \))
- \( x = -1 \) (for \( |x+1| \))
- \( x = 1 \) (for \( |x-1| \))

These points divide the interval \([-3, 2]\) into segments:
1. \( [-3, -2] \)
2. \( [-2, -1] \)
3. \( [-1, 1] \)
4. \( [1, 2] \)

### Step 2: Evaluate the integral on each segment

#### Segment 1: \( x \in [-3, -2] \)
For \( x < -2 \):
- \( |x+1| = -(x+1) = -x - 1 \)
- \( |x+2| = -(x+2) = -x - 2 \)
- \( |x-1| = -(x-1) = -x + 1 \)

Thus,
\[ |x+1| + |x+2| + |x-1| = (-x - 1) + (-x - 2) + (-x + 1) = -3x - 2 \]
The integral becomes:
\[ I_1 = \int_{-3}^{-2} (-3x - 2) \, dx \]

#### Segment 2: \( x \in [-2, -1] \)
For \( -2 \leq x < -1 \):
- \( |x+1| = -(x+1) = -x - 1 \)
- \( |x+2| = x + 2 \)
- \( |x-1| = -(x-1) = -x + 1 \)

Thus,
\[ |x+1| + |x+2| + |x-1| = (-x - 1) + (x + 2) + (-x + 1) = -x + 2 \]
The integral becomes:
\[ I_2 = \int_{-2}^{-1} (-x + 2) \, dx \]

#### Segment 3: \( x \in [-1, 1] \)
For \( -1 \leq x < 1 \):
- \( |x+1| = x + 1 \)
- \( |x+2| = x + 2 \)
- \( |x-1| = -(x-1) = -x + 1 \)

Thus,
\[ |x+1| + |x+2| + |x-1| = (x + 1) + (x + 2) + (-x + 1) = x + 4 \]
The integral becomes:
\[ I_3 = \int_{-1}^{1} (x + 4) \, dx \]

#### Segment 4: \( x \in [1, 2] \)
For \( x \geq 1 \):
- \( |x+1| = x + 1 \)
- \( |x+2| = x + 2 \)
- \( |x-1| = x - 1 \)

Thus,
\[ |x+1| + |x+2| + |x-1| = (x + 1) + (x + 2) + (x - 1) = 3x + 2 \]
The integral becomes:
\[ I_4 = \int_{1}^{2} (3x + 2) \, dx \]

### Step 3: Calculate each integral

1. **Calculate \( I_1 \)**:
\[ I_1 = \int_{-3}^{-2} (-3x - 2) \, dx = \left[-\frac{3}{2}x^2 - 2x\right]_{-3}^{-2} \]
\[ = \left[-\frac{3}{2}(-2)^2 - 2(-2)\right] - \left[-\frac{3}{2}(-3)^2 - 2(-3)\right] \]
\[ = \left[-6 + 4\right] - \left[-\frac{27}{2} + 6\right] = -2 + \frac{27}{2} - 6 = \frac{27 - 4}{2} = \frac{23}{2} \]

2. **Calculate \( I_2 \)**:
\[ I_2 = \int_{-2}^{-1} (-x + 2) \, dx = \left[-\frac{1}{2}x^2 + 2x\right]_{-2}^{-1} \]
\[ = \left[-\frac{1}{2}(-1)^2 + 2(-1)\right] - \left[-\frac{1}{2}(-2)^2 + 2(-2)\right] \]
\[ = \left[-\frac{1}{2} - 2\right] - \left[-2 + 4\right] = -\frac{5}{2} + 2 = -\frac{1}{2} \]

3. **Calculate \( I_3 \)**:
\[ I_3 = \int_{-1}^{1} (x + 4) \, dx = \left[\frac{1}{2}x^2 + 4x\right]_{-1}^{1} \]
\[ = \left[\frac{1}{2}(1)^2 + 4(1)\right] - \left[\frac{1}{2}(-1)^2 + 4(-1)\right] \]
\[ = \left[\frac{1}{2} + 4\right] - \left[\frac{1}{2} - 4\right] = \frac{9}{2} + \frac{7}{2} = \frac{16}{2} = 8 \]

4. **Calculate \( I_4 \)**:
\[ I_4 = \int_{1}^{2} (3x + 2) \, dx = \left[\frac{3}{2}x^2 + 2x\right]_{1}^{2} \]
\[ = \left[\frac{3}{2}(2)^2 + 2(2)\right] - \left[\frac{3}{2}(1)^2 + 2(1)\right] \]
\[ = \left[6 + 4\right] - \left[\frac{3}{2} + 2\right] = 10 - \frac{7}{2} = \frac{20 - 7}{2} = \frac{13}{2} \]

### Step 4: Combine the results
Now, we sum all the integrals:
\[ I = I_1 + I_2 + I_3 + I_4 = \frac{23}{2} - \frac{1}{2} + 8 + \frac{13}{2} \]
\[ = \frac{23 - 1 + 16 + 13}{2} = \frac{51}{2} \]

### Final Answer
Thus, the value of the integral \( I \) is:
\[ \boxed{\frac{51}{2}} \]