To find the future value (FV) of a continuous income stream at a continuously compounded interest rate, we use the formula:

$FV = \int_{0}^{T} f(t) e^{r(T-t)} \, dt$

Where:

• $f(t) = 1350 e^{-0.02t}$ is the rate of flow.
• $r = 4.75\% = 0.0475$ is the annual interest rate.
• $T = 6$ is the time period in years.

Step 1: Substitute the values into the formula:

$FV = \int_{0}^{6} 1350 e^{-0.02t} e^{0.0475(6-t)} \, dt$

Simplify the exponents: $FV = \int_{0}^{6} 1350 e^{0.285 - 0.02t - 0.0475t} \, dt$

Combine the exponents: $FV = \int_{0}^{6} 1350 e^{0.285 - 0.0675t} \, dt$

Step 2: Factor constants out of the integral:

$FV = 1350 e^{0.285} \int_{0}^{6} e^{-0.0675t} \, dt$

Step 3: Solve the integral:

The integral of $e^{kt}$ is: $\int e^{kt} \, dt = \frac{1}{k} e^{kt} + C$

Here, $k = -0.0675$, so: $\int e^{-0.0675t} \, dt = \frac{1}{-0.0675} e^{-0.0675t}$

Apply the limits $t = 0$ to $t = 6$: $\int_{0}^{6} e^{-0.0675t} \, dt = \left[\frac{1}{-0.0675} e^{-0.0675t} \right]_{0}^{6}$

$= \frac{1}{-0.0675} \left(e^{-0.0675(6)} - e^{-0.0675(0)}\right)$

$= \frac{1}{-0.0675} \left(e^{-0.405} - 1\right)$

Step 4: Combine the results:

$FV = 1350 e^{0.285} \cdot \frac{1}{-0.0675} \left(e^{-0.405} - 1\right)$

Numerically calculate each term:

• $e^{0.285} \approx 1.329$
• $e^{-0.405} \approx 0.667$
• $\frac{1}{-0.0675} \approx -14.815$

Substitute: $FV \approx 1350 \cdot 1.329 \cdot (-14.815) \cdot (0.667 - 1)$

$FV \approx 1350 \cdot 1.329 \cdot (-14.815) \cdot -0.333$

$FV \approx 1350 \cdot 1.329 \cdot 4.932$

$FV \approx 1350 \cdot 6.554$

$FV \approx 8858$

Final Answer:

The future value is approximately $8,858.

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Do you want more details or have questions?

Related Questions:

1. How does continuous compounding differ from regular compounding?
2. What does the integral represent in this context?
3. How would the future value change if the interest rate increased?
4. What if the time period was extended to 10 years?
5. How does the decay factor ($-0.02t$) of $f(t)$ affect the result?

Tip:

When dealing with continuous income streams, always combine exponents carefully before integration to simplify calculations.