Using integration, find the area of the region bounded by the lines `y=1+abs(x+1), x=-2, x=3 and y=0`.

Answer

To find the area of the region bounded by the lines \( y = 1 + |x + 1| \), \( x = -2 \), \( x = 3 \), and \( y = 0 \), we will follow these steps:

### Step 1: Understand the function
The function \( y = 1 + |x + 1| \) can be split into two cases based on the value of \( x \):
- For \( x \geq -1 \), \( |x + 1| = x + 1 \) so \( y = 1 + (x + 1) = x + 2 \).
- For \( x < -1 \), \( |x + 1| = -(x + 1) \) so \( y = 1 - (x + 1) = -x \).

### Step 2: Identify the boundaries
We need to find the area between the curves and the lines \( x = -2 \), \( x = 3 \), and the x-axis \( y = 0 \). The points of intersection will help us determine the limits of integration.

### Step 3: Find points of intersection
1. **For \( x = -2 \)**:
- \( y = 1 + |(-2) + 1| = 1 + | -1 | = 1 + 1 = 2 \)
- Point: \( (-2, 2) \)

2. **For \( x = -1 \)**:
- \( y = 1 + |(-1) + 1| = 1 + |0| = 1 \)
- Point: \( (-1, 1) \)

3. **For \( x = 3 \)**:
- \( y = 1 + |3 + 1| = 1 + 4 = 5 \)
- Point: \( (3, 5) \)

### Step 4: Set up the integrals
The area can be divided into two parts:
- Area \( A_1 \) from \( x = -2 \) to \( x = -1 \) under the curve \( y = -x \).
- Area \( A_2 \) from \( x = -1 \) to \( x = 3 \) under the curve \( y = x + 2 \).

Thus, the total area \( A \) is given by:
\[ A = A_1 + A_2 = \int_{-2}^{-1} (-x) \, dx + \int_{-1}^{3} (x + 2) \, dx \]

### Step 5: Calculate \( A_1 \)
\[ A_1 = \int_{-2}^{-1} (-x) \, dx = -\left[ \frac{x^2}{2} \right]_{-2}^{-1} = -\left( \frac{(-1)^2}{2} - \frac{(-2)^2}{2} \right) = -\left( \frac{1}{2} - 2 \right) = -\left( \frac{1}{2} - \frac{4}{2} \right) = -\left( -\frac{3}{2} \right) = \frac{3}{2} \]

### Step 6: Calculate \( A_2 \)
\[ A_2 = \int_{-1}^{3} (x + 2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{-1}^{3} = \left( \frac{3^2}{2} + 2 \cdot 3 \right) - \left( \frac{(-1)^2}{2} + 2 \cdot (-1) \right) \]
\[ = \left( \frac{9}{2} + 6 \right) - \left( \frac{1}{2} - 2 \right) = \left( \frac{9}{2} + \frac{12}{2} \right) - \left( \frac{1}{2} - \frac{4}{2} \right) \]
\[ = \frac{21}{2} - \left( -\frac{3}{2} \right) = \frac{21}{2} + \frac{3}{2} = \frac{24}{2} = 12 \]

### Step 7: Combine the areas
\[ A = A_1 + A_2 = \frac{3}{2} + 12 = \frac{3}{2} + \frac{24}{2} = \frac{27}{2} = 13.5 \]

### Final Answer
The area of the region bounded by the lines is \( 13.5 \) square units.