Using intergration, find the area of the region bounded by the lines `y=|x+1|,x= -3, x=1` and X-axis.

Answer

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

### Step 1: Understand the Absolute Value Function
The function \( y = |x + 1| \) can be expressed as:
- \( y = -(x + 1) \) for \( x < -1 \)
- \( y = x + 1 \) for \( x \geq -1 \)

### Step 2: Identify the Points of Intersection
We need to find the points where the lines intersect the x-axis:
- For \( y = 0 \):
- \( |x + 1| = 0 \) gives \( x = -1 \).

The lines \( x = -3 \) and \( x = 1 \) will also be considered.

### Step 3: Sketch the Graph
Sketching the graph helps visualize the area:
- The line \( y = -(x + 1) \) will be valid from \( x = -3 \) to \( x = -1 \).
- The line \( y = x + 1 \) will be valid from \( x = -1 \) to \( x = 1 \).

### Step 4: Set Up the Integrals
We will calculate the area in two parts:
1. Area from \( x = -3 \) to \( x = -1 \) under the curve \( y = -(x + 1) \).
2. Area from \( x = -1 \) to \( x = 1 \) under the curve \( y = x + 1 \).

#### Area 1:
\[ A_1 = \int_{-3}^{-1} (-(x + 1)) \, dx \]

#### Area 2:
\[ A_2 = \int_{-1}^{1} (x + 1) \, dx \]

### Step 5: Calculate Area 1
\[ A_1 = \int_{-3}^{-1} (-(x + 1)) \, dx = \int_{-3}^{-1} (-x - 1) \, dx \]
Calculating the integral:
\[ = \left[-\frac{x^2}{2} - x \right]_{-3}^{-1} \]
Calculating the limits:
\[ = \left[-\frac{(-1)^2}{2} - (-1)\right] - \left[-\frac{(-3)^2}{2} - (-3)\right] \]
\[ = \left[-\frac{1}{2} + 1\right] - \left[-\frac{9}{2} + 3\right] \]
\[ = \left[\frac{1}{2}\right] - \left[-\frac{3}{2}\right] = \frac{1}{2} + \frac{3}{2} = 2 \]

### Step 6: Calculate Area 2
\[ A_2 = \int_{-1}^{1} (x + 1) \, dx \]
Calculating the integral:
\[ = \left[\frac{x^2}{2} + x\right]_{-1}^{1} \]
Calculating the limits:
\[ = \left[\frac{(1)^2}{2} + (1)\right] - \left[\frac{(-1)^2}{2} + (-1)\right] \]
\[ = \left[\frac{1}{2} + 1\right] - \left[\frac{1}{2} - 1\right] \]
\[ = \left[\frac{3}{2}\right] - \left[-\frac{1}{2}\right] = \frac{3}{2} + \frac{1}{2} = 2 \]

### Step 7: Total Area
Now, we sum the areas:
\[ \text{Total Area} = A_1 + A_2 = 2 + 2 = 4 \]

### Final Answer
The area of the region bounded by the lines is \( 4 \) square units.

---