Problem Setup

The rate of water flow into the pool is constant at $5 \, \text{gallons per minute}$. Since the water flow rate is constant, the total amount of water added over time can be calculated by integrating the rate function with respect to time.

Integral Representation

Let $t$ represent time in minutes. The flow rate function is $r(t) = 5 \, \text{gallons per minute}$. The integral to represent the total amount of water added over $2$ hours is:

$\text{Total Water} = \int_{0}^{120} r(t) \, dt = \int_{0}^{120} 5 \, dt$

(Note: $2 \, \text{hours} = 120 \, \text{minutes}$.)

Evaluating the Integral

$\int_{0}^{120} 5 \, dt = 5 \int_{0}^{120} 1 \, dt = 5 \left[ t \right]_{0}^{120}$

$= 5 \cdot (120 - 0) = 600 \, \text{gallons}$

Final Answer

The total amount of water in the pool after $2 \, \text{hours}$ is:

$\boxed{600 \, \text{gallons}}$

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Would you like to explore a related concept or see alternative methods? Here are some related questions:

1. How would the calculation change if the flow rate were not constant?
2. What is the general formula for calculating total water added if $r(t)$ is a function of time?
3. How would the calculation change if the pool had an initial amount of water in it?
4. What if the flow rate was given in liters per second? How would the conversion work?
5. Can you derive a relationship for the water volume if the pool starts leaking at a constant rate?

Tip: When working with problems involving rates, always confirm that the units of time and rate are compatible for integration.