The given equation is $C(t) = 2 + t^{1.5}$, which represents the rate of increase in the maintenance cost of power drills over time. To find the total maintenance cost during the first five weeks, we need to integrate this function from $t = 0$ to $t = 5$.

The integral of $C(t)$ will give us the total maintenance cost over the given time period:

$\int_0^5 (2 + t^{1.5}) \, dt$

Step 1: Integrating the function

We can break this into two parts:

$\int_0^5 (2) \, dt + \int_0^5 (t^{1.5}) \, dt$

Part 1: Integrating $2$

$\int_0^5 2 \, dt = 2t \Big|_0^5 = 2(5) - 2(0) = 10$

Part 2: Integrating $t^{1.5}$

The integral of $t^{1.5}$ is:

$\int_0^5 t^{1.5} \, dt = \frac{t^{2.5}}{2.5} \Big|_0^5 = \frac{5^{2.5}}{2.5} - \frac{0^{2.5}}{2.5}$

Now, calculate $5^{2.5}$:

$5^{2.5} = 5 \times \sqrt{25} = 5 \times 5 = 25$

So:

$\frac{5^{2.5}}{2.5} = \frac{25}{2.5} = 10$

Step 2: Adding the results

Now, adding the two parts together:

$10 + 10 = 20$

Final Answer:

So, the total maintenance cost during the first five weeks is approximately 20.

Therefore, the closest answer is:

$\boxed{b. \ 20.5}$

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Do you have any questions or would you like further details?

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Here are 5 related questions to explore further:

1. How do we interpret the rate of increase function in terms of company operations?
2. How would the total cost change if the company operated for 10 weeks instead of 5?
3. What happens to the maintenance cost function if we change the exponent on $t$?
4. How can integration be used to solve real-world problems involving rates of change?
5. Can you determine the average maintenance cost over the first five weeks?

Tip: When integrating power functions, remember that the exponent increases by one and you divide by the new exponent.