To find the area between the curves, we calculate the integral of the difference between the two functions over the interval given. Here's the step-by-step solution:

Functions and Interval:

• $y_1 = 2e^{3x}$
• $y_2 = e^{3x} + e^6$
• Interval: $x \in [0, 3]$

The formula for the area between the curves is: $\text{Area} = \int_{0}^{3} |y_1 - y_2| \, dx$

Step 1: Calculate $y_1 - y_2$

$y_1 - y_2 = 2e^{3x} - (e^{3x} + e^6) = e^{3x} - e^6$

Step 2: Set Up the Integral

$\text{Area} = \int_{0}^{3} (e^{3x} - e^6) \, dx$

Step 3: Integrate

• The integral of $e^{3x}$ is $\frac{1}{3}e^{3x}$.
• The integral of $e^6$ is $e^6x$ (since $e^6$ is constant).

$\int (e^{3x} - e^6) \, dx = \frac{1}{3}e^{3x} - e^6x$

Step 4: Evaluate the Definite Integral

Substitute the limits $x = 0$ and $x = 3$: $\text{Area} = \left[\frac{1}{3}e^{3x} - e^6x \right]_{0}^{3}$

At $x = 3$: $\frac{1}{3}e^{9} - e^6 \cdot 3$

At $x = 0$: $\frac{1}{3}e^{0} - e^6 \cdot 0 = \frac{1}{3}$

Subtract the results: $\text{Area} = \left(\frac{1}{3}e^{9} - 3e^6\right) - \frac{1}{3}$ $\text{Area} = \frac{1}{3}e^{9} - 3e^6 - \frac{1}{3}$

Step 5: Numerical Approximation

Using approximate values:

• $e^3 \approx 20.0855$, so $e^6 \approx 403.4288$, and $e^9 \approx 8103.0839$.

Substitute these into the expression: $\text{Area} \approx \frac{1}{3}(8103.0839) - 3(403.4288) - \frac{1}{3}$ $\text{Area} \approx 2701.028 - 1210.286 - 0.333$ $\text{Area} \approx 1490.409$

Final Answer:

The area between the curves is approximately 1490 (rounded to the nearest whole number).

Would you like a detailed explanation for any specific step? Here are some follow-up questions:

1. How do we handle integrals when functions cross each other within the interval?
2. What are the applications of finding the area between curves?
3. Why is $e^x$ significant in exponential growth problems?
4. Can you explain how to handle definite integrals numerically using a calculator?
5. How does the integral's value change with a different interval?

Tip: Always check if the curves intersect within the given interval to confirm the proper limits of integration.