To determine the population after one hour, we first calculate the change in the population using the integral and then add this to the initial population.

Step 1: Define the rate of growth

The rate of growth of the population is given as:

$\frac{dP}{dt} = 1000 \cdot 7^t$

Step 2: Set up the integral for the change in population

The change in the population after one hour is represented by the definite integral:

$\Delta P = \int_0^1 1000 \cdot 7^t \, dt$

Step 3: Solve the integral

The integral of $1000 \cdot 7^t$ is calculated as follows:

$\int 1000 \cdot 7^t \, dt = \frac{1000}{\ln(7)} \cdot 7^t$

Evaluate this from $t = 0$ to $t = 1$:

$\Delta P = \left[ \frac{1000}{\ln(7)} \cdot 7^t \right]_0^1 = \frac{1000}{\ln(7)} \cdot 7^1 - \frac{1000}{\ln(7)} \cdot 7^0$

Simplify:

$\Delta P = \frac{1000}{\ln(7)} \cdot 7 - \frac{1000}{\ln(7)}$

$\Delta P = \frac{1000}{\ln(7)} \cdot (7 - 1)$

Step 4: Approximate numerical values

The natural logarithm of 7 is approximately $\ln(7) \approx 1.94591$. Thus:

$\Delta P = \frac{1000}{1.94591} \cdot 6 \approx 3082.4$

Step 5: Compute the total population after one hour

The initial population is $6000$. Adding the change in population:

$P(1) = 6000 + 3082.4 \approx 9082$

Final Answer:

The population after one hour is approximately 9082.

Would you like further clarification on any of these steps? Here are related questions you might explore:

1. What is the significance of using definite integrals for growth rates?
2. How does the base $7$ affect the population growth rate?
3. How would the answer change if the growth rate formula was altered?
4. Why is the natural logarithm ($\ln$) used in solving the integral?
5. How can this method be applied to other real-world population models?

Tip: When solving problems involving exponential growth, always double-check the integration constant and bounds for accurate calculations.