To solve this problem, we need to calculate the present value of an investment that generates continuous income at a rate of $f(t) = 2300 + 150t$ dollars per year, given a continuous compounding interest rate of 5%.

1. Formulate the Present Value Integral:

   The present value $PV$ of a continuous income stream $f(t)$ discounted at a continuous interest rate $r$ is given by: $PV = \int_{0}^{\infty} f(t) e^{-rt} \, dt$ Here, $f(t) = 2300 + 150t$ and $r = 0.05$.

2. Substitute $f(t)$ into the Integral: $PV = \int_{0}^{\infty} (2300 + 150t) e^{-0.05t} \, dt$

3. Split the Integral: $PV = \int_{0}^{\infty} 2300 e^{-0.05t} \, dt + \int_{0}^{\infty} 150t e^{-0.05t} \, dt$

4. Solve Each Integral Separately:

   For the first integral: $\int_{0}^{\infty} 2300 e^{-0.05t} \, dt = 2300 \int_{0}^{\infty} e^{-0.05t} \, dt$ [ \int_{0}^{\infty} e^{-0.05t} , dt = \left[ \frac{e^{-0.05t}}{-0.05} \right]_{0}^{\infty} = \left[ 0 - \frac{1}{-0.05} \right] = \frac{1}{0.05} = 20 ] Therefore: $2300 \times 20 = 46000$

   For the second integral: $\int_{0}^{\infty} 150t e^{-0.05t} \, dt$ This is a standard integral of the form $\int_{0}^{\infty} t e^{-at} \, dt = \frac{1}{a^2}$: $\int_{0}^{\infty} 150t e^{-0.05t} \, dt = 150 \int_{0}^{\infty} t e^{-0.05t} \, dt = 150 \times \frac{1}{0.05^2} = 150 \times 400 = 60000$

5. Add the Two Results: $PV = 46000 + 60000 = 106000$

Therefore, the present value of the investment is $106,000.

Do you want details or have any questions? Here are some questions you might ask:

1. How do you calculate the present value of continuous cash flows?
2. Can you explain the continuous compounding interest formula?
3. What is the significance of the exponential discount factor $e^{-rt}$?
4. How do you handle integrals involving time-dependent cash flows?
5. Can you show another example with different functions or interest rates?

Tip: When solving problems involving continuous cash flows, breaking down the integral into manageable parts can simplify the calculation.