Find the total area between the function f(x)=2x and the x-axis over the interval [-3,3]

To find the total area between the function \( f(x) = 2x \) and the x-axis over the interval \([-3, 3]\), we need to compute the definite integral of \( f(x) \) over this interval. 

### Step 1: Determine the function and interval
The function is \( f(x) = 2x \) and the interval is \([-3, 3]\).

### Step 2: Identify the points where \( f(x) \) intersects the x-axis
The function \( f(x) = 2x \) intersects the x-axis where \( f(x) = 0 \). Solving for \( x \):
\[
2x = 0 \implies x = 0
\]
So, the function intersects the x-axis at \( x = 0 \).

### Step 3: Compute the area
Since the function changes sign at \( x = 0 \), we need to break the integral into two parts: one for \([-3, 0]\) and one for \([0, 3]\).

1. **For the interval \([-3, 0]\)**:
\[
\text{Area}_1 = \int_{-3}^{0} 2x \, dx
\]

2. **For the interval \([0, 3]\)**:
\[
\text{Area}_2 = \int_{0}^{3} 2x \, dx
\]

### Step 4: Evaluate the integrals
1. **Evaluate the first integral**:
\[
\text{Area}_1 = \int_{-3}^{0} 2x \, dx = \left[ x^2 \right]_{-3}^{0} = 0^2 - (-3)^2 = 0 - 9 = -9
\]
Since area cannot be negative, we take the absolute value: \( \text{Area}_1 = 9 \).

2. **Evaluate the second integral**:
\[
\text{Area}_2 = \int_{0}^{3} 2x \, dx = \left[ x^2 \right]_{0}^{3} = 3^2 - 0^2 = 9 - 0 = 9
\]

### Step 5: Add the areas
The total area between the function and the x-axis over the interval \([-3, 3]\) is:
\[
\text{Total Area} = \text{Area}_1 + \text{Area}_2 = 9 + 9 = 18
\]

Thus, the total area between the function \( f(x) = 2x \) and the x-axis over the interval \([-3, 3]\) is **18 square units**.

Would you like more details on any part of this solution, or do you have any other questions?

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Here are 8 related questions you might consider:

1. How would the area change if the function was \( f(x) = 3x \) instead of \( 2x \)?
2. What is the geometric interpretation of the area calculated?
3. How do you calculate the area between a quadratic function and the x-axis?
4. How would the integral change if the interval was \([-4, 2]\)?
5. What is the significance of the absolute value in determining the area?
6. Can this method be applied to find the area between \( f(x) = x^2 \) and the x-axis?
7. How would you find the area between two different linear functions?
8. How do you interpret the negative area in definite integrals?

**Tip:** When calculating the area under a curve, always consider breaking the interval at points where the function crosses the x-axis. This ensures that you properly account for areas below and above the axis.

Learn how to calculate the total area between the function f(x)=2x and the x-axis over the interval [-3, 3] using definite integrals. Understand the concept of breaking the interval and evaluating integrals for different segments. Suitable for advanced high school students and college-level calculus enthusiasts.

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