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.