Question

You are given the rate of investment \(d l / d t\). Find the capital accumulation over a five-year period by evaluating the definite integral Capital accumulation \(=\int_{0}^{5} \frac{d l}{d t} d t\) where \(t\) is the time in years. $$ \frac{d I}{d t}=500 \sqrt{t+1} $$

Short answer

The total capital accumulation over a five-year period is calculated by integrating the rate of investment function from 0 to 5 years, which gives approximate value of 6081.02 unit of capital.

Step-by-step solution

Identify the function to integrate

The function to integrate is the rate of investment, which is \(d I / d t = 500 \sqrt{t+1}\)

Set up the integral

The integral of the rate of investment function over a given period of time gives the total capital accumulation. Therefore, calculate the definite integral from 0 to 5 years: \(\int_{0}^{5} 500 \sqrt{t+1} dt\).

Evaluate the integral

Solving this integral involves a basic power rule for integration, \(∫x^n dx = x^{n+1} / (n+1) + C\), where \(C\) is the constant of integration. Apply this rule to solve the integral: \(500 * \frac{2}{3} * ((t+1)^{3/2})\Big|_0^{5}\).

Calculate total capital

Substitute \(t=5\) and \(t=0\) in the expression to find the accumulated capital from years 0 to 5. Capital accumulation = \(500 * \frac{2}{3} * ((5+1)^{3/2}) - 500 * \frac{2}{3} * ((0+1)^{3/2})\).

Key concepts

Definite Integral

Understanding the definite integral is crucial when studying calculus, especially in the context of real-world applications like capital accumulation. In essence, the definite integral is used to calculate the total accumulation of a quantity over a specific interval. Think of it as adding up an infinite number of infinitesimally small quantities between a start point and an end point.

For example, when you are given the rate of investment \( \frac{dI}{dt} \), which tells you how much investment is changing over time, you can use the definite integral to find the total change in investment over a period. In the provided exercise, we're interested in finding out the total capital accumulation over five years. To do this, we set up an integral from 0 to 5 years with the rate of investment as the integrand, which is mathematically represented by \( \[ \int_{0}^{5} \frac{dI}{dt} dt \] \).

This process is akin to adding up the money invested at each infinitesimal moment from the start of the first year to the end of the fifth year. The result is the total amount of capital that has been accumulated. The definite integral provides a powerful tool to sum up continuous rates of change and is widely used in economics, physics, and engineering.

Rate of Investment

In financial terms, the rate of investment is a measure of how quickly investment is being made into a particular asset or portfolio over time. In the context of calculus, it's a function that describes how investment is accumulating at each moment in time. The rate of investment may change from moment to moment, which is why we use calculus to understand the overall effect over a period.

Intuitively, you might think of the rate of investment as how much money per year is put into an investment. In the exercise's equation \( \frac{dI}{dt} = 500 \sqrt{t+1} \), the rate of investment varies depending on time \( t \), which requires the use of calculus to fully grasp the total amount of investment over time. This function implies that the investment rate increases as time goes on, since the value of the square root will get larger as \( t \), the time in years, increases.

When you work with problems involving the rate of investment, it's important to understand both the conceptual and mathematical side of it. Conceptually, why the rate of investment changes as it does, and mathematically, how to work with this rate to calculate total investments using integration.

Power Rule Integration

When it comes to integrating polynomial functions, the power rule for integration is one of the fundamental techniques in calculus. This rule states that the integral of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} \) plus a constant of integration \( C \), symbolically expressed as \( ∫x^n dx = \frac{x^{n+1}}{n+1} + C \).

The exercise at hand requires applying this rule. However, since we have a square root, we first rewrite it in terms of powers: \( \sqrt{t+1} = (t+1)^{1/2} \). Now the power rule integration can be applied to the function \( 500(t+1)^{1/2} \), resulting in \( 500 * \frac{2}{3} * (t+1)^{3/2} \) from the application of the power rule \( ∫(t+1)^{1/2} dt = \frac{2}{3}(t+1)^{3/2} \).

The constant of integration is omitted when evaluating a definite integral because it cancels out when taking the difference between the upper and lower bounds. Using the power rule simplifies the process of finding integrals of polynomials and is a staple in a student's calculus toolkit.