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Let's solve the given problem step-by-step.

### Problem Statement
The function $ f(t) = 1500 $ represents the rate of flow of money in dollars per year. Assume a 15-year period at 4% compounded continuously. We need to:
(a) Find the present value.
(b) Find the accumulated amount of money flow at $ T = 15 $.

### Solution
#### (a) Present Value
To find the present value, we use the following formula:
\[ P = \int_{0}^{15} f(t) e^{-rt} \, dt \]

Where:
- $ f(t) = 1500 $
- $ r = 0.04 $
- $ t $ is the time in years, ranging from 0 to 15

So, the integral becomes:
\[ P = \int_{0}^{15} 1500 e^{-0.04t} \, dt \]

We need to select the correct choice and fill in the integral. Looking at the provided options, the correct choice is:

\[
\text{Choice A:} \quad P = e^{0.04 \cdot \text{blank}} \int_{0}^{15} 1500 e^{-0.04 t} \, dt
\]

Fill in the blanks:
- The exponent in $ e^{0.04 \cdot \text{blank}} $ should be 0.
- The integral part is already correctly set up.

So, the correct setup for the integral is:

\[
P = \int_{0}^{15} 1500 e^{-0.04 t} \, dt
\]

Now, let's compute the integral:
\[
P = 1500 \int_{0}^{15} e^{-0.04 t} \, dt
\]

To solve the integral, we use the following formula for the integral of an exponential function:
\[
\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C
\]

Here, $ a = -0.04 $, so:
\[
P = 1500 \left[ \frac{e^{-0.04 t}}{-0.04} \right]_{0}^{15}
\]

Simplify the expression:
\[
P = 1500 \left( \frac{e^{-0.04 \cdot 15}}{-0.04} - \frac{e^{-0.04 \cdot 0}}{-0.04} \right)
\]

\[
P = 1500 \left( \frac{e^{-0.6}}{-0.04} - \frac{1}{-0.04} \right)
\]

\[
P = 1500 \left( \frac{1 - e^{-0.6}}{0.04} \right)
\]

\[
P = \frac{1500}{0.04} (1 - e^{-0.6})
\]

\[
P = 37500 (1 - e^{-0.6})
\]

Now we compute the numerical value:
\[
e^{-0.6} \approx 0.5488
\]

\[
P = 37500 (1 - 0.5488)
\]

\[
P = 37500 \times 0.4512
\]

\[
P \approx 16920
\]

So, the present value is approximately \$16,920.00.

### (b) Accumulated Amount of Money Flow at $ T = 15 $
The accumulated amount $ A $ is given by the formula for continuous compounding:
\[ A = P e^{rT} \]

Using the present value $ P \approx 16920 $, $ r = 0.04 $, and $ T = 15 $:
\[
A = 16920 \times e^{0.04 \times 15}
\]

\[
A = 16920 \times e^{0.6}
\]

\[
e^{0.6} \approx 1.8221
\]

\[
A \approx 16920 \times 1.8221
\]

\[
A \approx 30839.82
\]

So, the accumulated amount of money flow at $ T = 15 $ is approximately \$30,839.82.

### Summary
(a) The present value is approximately \$16,920.00.
(b) The accumulated amount of money flow at $ T = 15 $ is approximately \$30,839.82.

Would you like further details or have any questions? Here are some additional questions for you to explore:

1. How does the present value change if the interest rate changes to 5%?
2. What is the accumulated amount if the time period is extended to 20 years?
3. How would the integral change if the rate of flow of money $ f(t) $ was a function of time?
4. Can you derive the formula for the present value using integration by parts?
5. How does continuous compounding differ from discrete compounding in terms of present value calculations?

**Tip:** Always verify the calculations by breaking them into smaller steps and using known approximations for exponential functions.

Learn how to calculate the present value and accumulated amount of money flow over 15 years with continuous compounding. This detailed solution uses integral calculus and exponential functions, providing step-by-step calculations. Suitable for advanced high school or college-level students.

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