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To solve for the average population of the town between the years 2003 and 2009, we are given the population function:

\[
P(t) = 6503e^{0.035t}
\]

Where:
- \( P(t) \) represents the population at year \( t \), measured from the beginning of the year 2000.
- \( t = 3 \) corresponds to the beginning of 2003, and \( t = 9 \) corresponds to the beginning of 2009.

### Step 1: Average population formula
The formula for the average value of a continuous function \( f(t) \) over the interval \( [a, b] \) is:

\[
\text{Average Value} = \frac{1}{b - a} \int_{a}^{b} f(t)\, dt
\]

In our case, \( f(t) = P(t) = 6503e^{0.035t} \), and the interval is from \( t = 3 \) to \( t = 9 \). So, the average population is:

\[
\text{Average Population} = \frac{1}{9 - 3} \int_{3}^{9} 6503e^{0.035t}\, dt
\]

### Step 2: Compute the integral
We first compute the integral of \( 6503e^{0.035t} \). The integral of \( e^{kt} \) is:

\[
\int e^{kt} dt = \frac{1}{k} e^{kt}
\]

Thus:

\[
\int 6503e^{0.035t}\, dt = \frac{6503}{0.035} e^{0.035t}
\]

Now evaluate this at the bounds \( t = 9 \) and \( t = 3 \):

\[
\int_{3}^{9} 6503e^{0.035t}\, dt = \frac{6503}{0.035} \left( e^{0.035(9)} - e^{0.035(3)} \right)
\]

### Step 3: Simplify the expression
First, calculate the values inside the exponential:

- \( 0.035 \times 9 = 0.315 \), so \( e^{0.315} \approx 1.370 \)
- \( 0.035 \times 3 = 0.105 \), so \( e^{0.105} \approx 1.111 \)

Now substitute back into the integral:

\[
\int_{3}^{9} 6503e^{0.035t}\, dt \approx \frac{6503}{0.035} \left( 1.370 - 1.111 \right)
\]

\[
\int_{3}^{9} 6503e^{0.035t}\, dt \approx \frac{6503}{0.035} \times 0.259 \approx 185800.57
\]

### Step 4: Compute the average population
Now divide by \( 9 - 3 = 6 \):

\[
\text{Average Population} = \frac{185800.57}{6} \approx 30967
\]

### Final Answer:
The average population of the town between 2003 and 2009 is approximately **30,967** people.

Would you like more details on the integration process or other clarifications?

Here are some related questions for further exploration:

1. How would the result change if the growth rate was higher than 3.5%?
2. What would the average population be over a different time interval, say from 2005 to 2015?
3. Can you explain how the exponential growth function models real-world population growth?
4. How does the choice of the bounds \( t = 3 \) and \( t = 9 \) affect the result?
5. How would you calculate the total population change over the same interval?

**Tip:** When solving exponential growth problems, always pay attention to the rate of growth and ensure accurate calculations of the integral bounds.

The population of a town t years after the beginning of the year 2000 is given by P(t) = 6503e^(0.035t). Find the AVERAGE population of the town from the beginning of the year 2003 to the beginning of the year 2009. ROUND YOUR ANSWER TO THE NEAREST PERSON.

Learn how to find the average population of a town from 2003 to 2009 using the exponential growth function P(t) = 6503e^(0.035t). This solution involves applying calculus concepts, such as integration, and uses the formula for the average value of a continuous function. Step-by-step guidance included.

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