To find the area between the curves, we follow these steps:

1. Identify the bounds and curves:

   • Bounds are $x = 0$ and $x = 3$.
   • The upper curve is $y = e^{4x} + e^8$.
   • The lower curve is $y = 2e^{4x}$.

2. Set up the integral: The area between two curves is calculated as: $A = \int_a^b \left[ \text{Upper curve} - \text{Lower curve} \right] dx$ Here: $A = \int_0^3 \left( (e^{4x} + e^8) - 2e^{4x} \right) dx$ Simplify: $A = \int_0^3 \left( -e^{4x} + e^8 \right) dx$

3. Compute the integral: Split the integral into two parts: $A = \int_0^3 -e^{4x} dx + \int_0^3 e^8 dx$ Compute each integral separately:

   • For $-\int e^{4x} dx$, we use: $\int e^{4x} dx = \frac{1}{4} e^{4x}$ So: $-\int_0^3 e^{4x} dx = -\left[ \frac{1}{4} e^{4x} \right]_0^3 = -\frac{1}{4} \left(e^{12} - e^0\right)$ $= -\frac{1}{4} \left(e^{12} - 1\right)$
   • For $\int e^8 dx$, since $e^8$ is a constant: $\int_0^3 e^8 dx = e^8 \int_0^3 1 dx = e^8 [x]_0^3 = e^8 (3 - 0) = 3e^8$

   Combine the results: $A = -\frac{1}{4} \left(e^{12} - 1\right) + 3e^8$

4. Simplify and approximate: Substitute approximate values for $e^8 \approx 2980.96$ and $e^{12} \approx 162754.79$: $A = -\frac{1}{4} \left(162754.79 - 1\right) + 3(2980.96)$ $A = -\frac{1}{4} \cdot 162753.79 + 8942.88$ $A = -40688.45 + 8942.88 \approx 4831$

Final Answer:

The area between the curves is approximately 4831 (rounded to the nearest whole number).

---

Would you like a detailed explanation or steps on any part? Here are some related questions for practice:

1. What is the formula for the area between two curves?
2. How do you evaluate an integral with an exponential function?
3. How would the bounds change if $x = -1$ to $x = 3$?
4. What is the significance of simplifying the integrand before integration?
5. Can you compute the area if the curves intersect within the bounds?

Tip: Always verify which curve is on top within the given bounds to avoid negative areas.