91Ó°ÊÓ

A continuous income stream is a function that represents income being generated over time continuously. Unlike periodic payments, continuous income assumes that payments happen at every instant in an interval.

For example, imagine that a business generates revenue every second, without pause, rather than receiving money weekly or monthly.

The income stream is often modeled by a function, such as the given in the problem: \(R(t) = t^2\), where \(t\) represents the time at which the income is generated.

This function allows us to analyze and work with the income and its present value over any desired period.

Integration by Parts is a powerful technique used to solve more complex integrals. It's a little like the product rule for differentiation but in reverse.

The basic formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \]

In solving our problem, we applied integration by parts twice to evaluate \int_0^{40} t^2 e^{-0.07t} \, dt\.

First Integration by Parts:

• Choose \(u = t^2\) and \(dv = e^{-0.07t} dt\)
• Differentiate to get \(du = 2t \, dt\)
• Integrate to get \(v = -\frac{1}{0.07} e^{-0.07t}\)

Using the formula:
\[ \int_0^{40} t^2 e^{-0.07t} \, dt = \left[-\frac{t^2}{0.07} e^{-0.07t}\right]_0^{40} + \frac{2}{0.07} \int_0^{40} t e^{-0.07t} \, dt \]

Second Integration by Parts:

• Choose \(u = t\) and \(dv = e^{-0.07t} \, dt\)
• Differentiate to get \(du = dt\)
• Integrate to get \(v = -\frac{1}{0.07} e^{-0.07t}\)

Using the formula again:
\[ \int_0^{40} t e^{-0.07t} \, dt = \left[-\frac{t}{0.07} e^{-0.07t} \right]_0^{40} + \frac{1}{0.07} \int_0^{40} e^{-0.07t} \, dt \] By evaluating these, we can combine the results to achieve the desired integral.

Continuous compounding interest is a concept where the interest on an investment increases at every possible instant. This is different from yearly, quarterly, or even daily compounding.

With continuous compounding, the formula for interest is given by:
\[ A = Pe^{rt} \]

Where:

• \(A\) is the amount of money accumulated after n years, including interest
• \(P\) is the principal amount (the initial sum of money)
• \(r\) is the annual interest rate (decimal)
• \(t\) is the time the money is invested or borrowed for, in years
• \(e\) is the base of the natural logarithm

Applying this to our given problem, we use the continuous rate \(k=0.07\).

Therefore, this affects the accumulated present value formula:
\[ PV = \int_0^T R(t) e^{-kt} \, dt \]

Here, \(e^{-kt} \) represents the discount factor, accounting for continuous compounding interest.

It shows how much \($1\) today is worth in the future considering continuous growth at rate \ k \.