Sketch the graph of y=|x+4|. Using integration find the area of the region bounded by the curve y=|x+4| and x= -6 and x=0.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the function and its graph
The function given is \( y = |x + 4| \). The modulus function will change the behavior of the graph at the point where the expression inside the modulus equals zero, which is at \( x = -4 \).

### Step 2: Break down the function
We can express the function as two separate linear functions:
- For \( x < -4 \): \( y = -(x + 4) = -x - 4 \)
- For \( x \geq -4 \): \( y = x + 4 \)

### Step 3: Sketch the graph
1. **For \( x < -4 \)**:
- The line \( y = -x - 4 \) has a slope of -1 and a y-intercept of -4.
- It will intersect the y-axis at \( (0, -4) \).

2. **For \( x \geq -4 \)**:
- The line \( y = x + 4 \) has a slope of 1 and a y-intercept of 4.
- It will intersect the y-axis at \( (0, 4) \).

3. **Points of interest**:
- The vertex of the graph is at \( (-4, 0) \).
- The graph will cross the x-axis at \( (-4, 0) \) and \( (0, 4) \).

### Step 4: Identify the area to be calculated
We need to find the area bounded by the curve \( y = |x + 4| \) and the lines \( x = -6 \) and \( x = 0 \).

### Step 5: Set up the integral for the area
We will calculate the area in two parts:
1. From \( x = -6 \) to \( x = -4 \), where \( y = -x - 4 \).
2. From \( x = -4 \) to \( x = 0 \), where \( y = x + 4 \).

The area \( A \) can be expressed as:
\[ A = \int_{-6}^{-4} (-x - 4) \, dx + \int_{-4}^{0} (x + 4) \, dx \]

### Step 6: Calculate the first integral
\[ \int_{-6}^{-4} (-x - 4) \, dx = \left[ -\frac{x^2}{2} - 4x \right]_{-6}^{-4} \]
Calculating the limits:
- At \( x = -4 \):
\[ -\frac{(-4)^2}{2} - 4(-4) = -8 + 16 = 8 \]
- At \( x = -6 \):
\[ -\frac{(-6)^2}{2} - 4(-6) = -18 + 24 = 6 \]
Thus, the first integral evaluates to:
\[ 8 - 6 = 2 \]

### Step 7: Calculate the second integral
\[ \int_{-4}^{0} (x + 4) \, dx = \left[ \frac{x^2}{2} + 4x \right]_{-4}^{0} \]
Calculating the limits:
- At \( x = 0 \):
\[ 0 + 0 = 0 \]
- At \( x = -4 \):
\[ \frac{(-4)^2}{2} + 4(-4) = 8 - 16 = -8 \]
Thus, the second integral evaluates to:
\[ 0 - (-8) = 8 \]

### Step 8: Add the areas
Now, add the two areas together:
\[ A = 2 + 8 = 10 \]

### Final Answer
The area of the region bounded by the curve \( y = |x + 4| \) and the lines \( x = -6 \) and \( x = 0 \) is \( 10 \) square units.

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