Using integration, find the area of the region bounded by the curve `y= 1+|x+1|` and lines `x =-3, x = 3, y=0`.

Answer

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

### Step 1: Analyze the curve \( y = 1 + |x + 1| \)

The modulus function \( |x + 1| \) can be expressed in piecewise form:

- For \( x < -1 \): \( |x + 1| = -(x + 1) = -x - 1 \)
- For \( x \geq -1 \): \( |x + 1| = x + 1 \)

Thus, we can rewrite the function \( y \) as:

\[ y = \begin{cases} 1 - x & \text{if } x < -1 \\ x + 2 & \text{if } x \geq -1 \end{cases} \]

### Step 2: Find the points of intersection with the x-axis

To find the area, we need to determine where the curve intersects the line \( y = 0 \):

1. For \( x < -1 \):
\[ 1 - x = 0 \implies x = 1 \quad (\text{not in this interval}) \]

2. For \( x \geq -1 \):
\[ x + 2 = 0 \implies x = -2 \quad (\text{valid in this interval}) \]

Thus, the curve intersects the x-axis at \( x = -2 \).

### Step 3: Set up the area calculation

The area we want to calculate is bounded by the lines \( x = -3 \), \( x = -2 \), \( x = 3 \), and \( y = 0 \). We can break this area into two parts:

1. From \( x = -3 \) to \( x = -2 \) under the line \( y = 1 - x \).
2. From \( x = -2 \) to \( x = 3 \) under the line \( y = x + 2 \).

### Step 4: Calculate the area from \( x = -3 \) to \( x = -2 \)

The area \( A_1 \) under the curve from \( x = -3 \) to \( x = -2 \):

\[ A_1 = \int_{-3}^{-2} (1 - x) \, dx \]

Calculating the integral:

\[ A_1 = \left[ x - \frac{x^2}{2} \right]_{-3}^{-2} = \left( -2 - \frac{(-2)^2}{2} \right) - \left( -3 - \frac{(-3)^2}{2} \right) \]

Calculating the values:

\[ = \left( -2 - 2 \right) - \left( -3 - \frac{9}{2} \right) = -4 - \left( -3 - 4.5 \right) = -4 + 7.5 = 3.5 \]

### Step 5: Calculate the area from \( x = -2 \) to \( x = 3 \)

The area \( A_2 \) under the curve from \( x = -2 \) to \( x = 3 \):

\[ A_2 = \int_{-2}^{3} (x + 2) \, dx \]

Calculating the integral:

\[ A_2 = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{3} = \left( \frac{3^2}{2} + 2(3) \right) - \left( \frac{(-2)^2}{2} + 2(-2) \right) \]

Calculating the values:

\[ = \left( \frac{9}{2} + 6 \right) - \left( 2 - 4 \right) = \left( 4.5 + 6 \right) - (-2) = 10.5 + 2 = 12.5 \]

### Step 6: Total Area

The total area \( A \) is the sum of \( A_1 \) and \( A_2 \):

\[ A = A_1 + A_2 = 3.5 + 12.5 = 16 \]

Thus, the area of the region bounded by the curve and the lines is \( \boxed{16} \) square units.