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The problem asks to find the area between the graph of \( f(x) = 4e^x - 3 \) and the x-axis over the interval \([-4, 2]\), using the definite integral. It also instructs to check if the graph crosses the x-axis within this interval.

### Step 1: Check if the graph crosses the x-axis
To determine if the graph crosses the x-axis, we need to find where \( f(x) = 0 \):
\[
4e^x - 3 = 0
\]
Solving for \(x\):
\[
4e^x = 3 \quad \Rightarrow \quad e^x = \frac{3}{4} \quad \Rightarrow \quad x = \ln\left(\frac{3}{4}\right)
\]
Using the natural logarithm:
\[
x \approx \ln(0.75) \approx -0.2877
\]
So, the function crosses the x-axis at \( x \approx -0.2877 \), which is within the interval \([-4, 2]\).

### Step 2: Setting up the integral
Because the function crosses the x-axis at \( x = -0.2877 \), we need to split the integral at this point. The area above the x-axis will be positive, and the area below the x-axis will be negative, so we handle these as two separate integrals.

\[
\text{Area} = \left| \int_{-4}^{-0.2877} (4e^x - 3) \, dx \right| + \int_{-0.2877}^{2} (4e^x - 3) \, dx
\]

### Step 3: Solving the integral
The integral of \( 4e^x - 3 \) is straightforward:
\[
\int (4e^x - 3) \, dx = 4e^x - 3x
\]

Now, let's evaluate each part:

#### (a) From \( x = -4 \) to \( x \approx -0.2877 \):
\[
\left| \int_{-4}^{-0.2877} (4e^x - 3) \, dx \right| = \left| \left[ 4e^x - 3x \right]_{-4}^{-0.2877} \right|
\]

#### (b) From \( x \approx -0.2877 \) to \( x = 2 \):
\[
\int_{-0.2877}^{2} (4e^x - 3) \, dx = \left[ 4e^x - 3x \right]_{-0.2877}^{2}
\]

Let me calculate these integrals.The total area between the graph of \( f(x) = 4e^x - 3 \) and the x-axis over the interval \([-4, 2]\) is approximately 27.90 square units.

Would you like more details on any specific part of the solution or further explanation?

Here are five follow-up questions for deeper understanding:
1. How does the natural logarithm help in solving \(4e^x - 3 = 0\)?
2. Why do we split the integral at the point where the function crosses the x-axis?
3. How would you find the area if the function didn’t cross the x-axis in the interval?
4. Can you generalize this method to any function \(f(x)\) that crosses the x-axis?
5. What happens to the area calculation if the function \(f(x)\) is negative over the entire interval?

**Tip:** When calculating areas between curves, always be cautious of where the function crosses the x-axis; split the integral accordingly to avoid negative areas.

Use the definite integral to find the area between the x-axis and f(x) over the indicated interval. Check first to see if the graph crosses the x-axis in the given interval. f(x) = 4e^x - 3; [-4, 2]

Learn how to use the definite integral to find the area between the x-axis and the curve f(x) = 4e^x - 3 over the interval [-4, 2]. This problem includes checking where the graph crosses the x-axis and splitting the integral accordingly.

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