Draw a rough sketch of the given curve `y=1+abs(x+1),x=-3, x=3, y=0` and find the area of the region bounded by them, using integration.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given function
The function given is \( y = 1 + |x + 1| \). We need to analyze this function based on the definition of the absolute value.

### Step 2: Break down the absolute value
The absolute value function can be split into two cases:
1. When \( x + 1 \geq 0 \) (i.e., \( x \geq -1 \)):
\[ y = 1 + (x + 1) = x + 2 \]
2. When \( x + 1 < 0 \) (i.e., \( x < -1 \)):
\[ y = 1 - (x + 1) = 1 - x - 1 = -x \]

### Step 3: Identify the boundaries
We are also given the vertical lines \( x = -3 \) and \( x = 3 \), and the horizontal line \( y = 0 \).

### Step 4: Sketch the curves
- For \( x < -1 \), the line \( y = -x \) is drawn. This line will intersect the x-axis at \( (0, 0) \).
- For \( x \geq -1 \), the line \( y = x + 2 \) will intersect the y-axis at \( (0, 2) \).
- The vertical lines at \( x = -3 \) and \( x = 3 \) will be drawn, and the horizontal line \( y = 0 \) will represent the x-axis.

### Step 5: Find the area between the curves
To find the area between these curves, we need to integrate the difference between the upper function and the lower function over the specified intervals.

#### Area Calculation
1. **From \( x = -3 \) to \( x = -1 \)**:
- The upper function is \( y = -x \) and the lower function is \( y = 0 \).
- The area \( A_1 \) is given by:
\[ A_1 = \int_{-3}^{-1} (-x - 0) \, dx = \int_{-3}^{-1} -x \, dx \]

2. **From \( x = -1 \) to \( x = 3 \)**:
- The upper function is \( y = x + 2 \) and the lower function is \( y = 0 \).
- The area \( A_2 \) is given by:
\[ A_2 = \int_{-1}^{3} (x + 2 - 0) \, dx = \int_{-1}^{3} (x + 2) \, dx \]

### Step 6: Calculate the integrals
1. **Calculate \( A_1 \)**:
\[ A_1 = \int_{-3}^{-1} -x \, dx = \left[-\frac{x^2}{2}\right]_{-3}^{-1} = \left[-\frac{(-1)^2}{2} - \left(-\frac{(-3)^2}{2}\right)\right] = \left[-\frac{1}{2} + \frac{9}{2}\right] = \frac{8}{2} = 4 \]

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

### Step 7: Total Area
The total area \( A \) is the sum of \( A_1 \) and \( A_2 \):
\[ A = A_1 + A_2 = 4 + 12 = 16 \]

### Final Answer
The area of the region bounded by the curves is \( \boxed{16} \) square units.