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Annual revenues in 1996-2006 from slot machines and video poker machines at the Mohegan Sun casino in Connecticut can be modeled by $$ R(t)=1,260-1,030 e^{-0.1 t} \text { million dollars } \quad(0 \leq t \leq 10) $$ where \(t\) is time in years since \(1996 .{ }^{14}\) Find an expression for the total revenue earned since 1996 , and hence estimate the total revenue earned over the period 1996-2006. (Round your answer to the nearest $$\$ 10$$ million.)

Short Answer

Expert verified
The total revenue earned over the period 1996-2006 is approximately \(\boxed{12,600\text{ million dollars}}\) when rounded to the nearest 10 million dollars.

Step by step solution

01

1. Given function

The given function for annual revenue is: $$ R(t) = 1,260 - 1,030 e^{-0.1 t}\text{ million dollars}\quad (0 \leq t \leq 10) $$
02

2. Finding an expression for the total revenue

To find the total revenue, we need to integrate R(t) with respect to t from 0 to 10: $$ Total\_Revenue = \int_{0}^{10} R(t) dt = \int_{0}^{10} (1,260 - 1,030 e^{-0.1 t}) dt $$
03

3. Integrating the function

Now we integrate the function with respect to t: $$ \begin{aligned} \int_{0}^{10} (1,260 - 1,030 e^{-0.1 t}) dt &= \int_{0}^{10} 1,260 dt - \int_{0}^{10} 1,030 e^{-0.1 t} dt\\ &= 1,260\int_{0}^{10} dt - 1,030 \int_{0}^{10} e^{-0.1t} dt \end {aligned} $$ To integrate the second term, we'll substitute a variable, \(u = -0.1t\). Therefore, \(-10du = dt\). Then, the integration becomes: $$ 1,260\int_{0}^{10} dt - 1,030 \int_{0}^{10} e^{-0.1t} dt = 1,260t\Big|_0^{10} - 1,030\int_{-10^{-5}} e^u (-10)du $$
04

4. Evaluating the integrals

Now we evaluate the integrals: $$ \begin{aligned} 1,260t\Big|_0^{10} -1,030\int_{-10^{-5}} e^u (-10)du &= 1,260(10) - 1,030(-10)\int_{-10^{-5}} e^u du\\ &= 12,600 - 10,300\left[ \int_{-10^{-5}} e^u du \right]\\ &= 12,600 - 10,300 (e^u)\Big|_{-10^{-5}} \\ \end{aligned} $$ Calculate the value of the definite integral: $$ \begin{aligned} 12,600 - 10,300 (e^u)\Big|_{-10^{-5}} &= 12,600 - 10,300(e^{10^{-5}} - e^0) \\ &= 12,600 - 10,300(1.0001 - 1) \\ \end{aligned} $$
05

5. Estimating the total revenue

Now we will estimate the total revenue and round it to the nearest $10 million: $$ \begin{aligned} 12,600 - 10,300(1.0001- 1) &= 12,600 - 10,300(0.0001)=12,600-1.03\\ &\approx 12,599 \end{aligned} $$ So the estimated total revenue earned over the period 1996-2006 is approximately \(12,599\) million, or rounded to the nearest \(10\) million, it is \(\boxed{12,600\text{ million dollars}}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
Integration is a fundamental concept in calculus that represents the accumulation of quantities, such as area under a curve, total distance travelled, or, as in our example, the total revenue earned over a period of time. The \textbf{definite integral} is specifically used when we want to calculate the total accumulation between two points, or in the case of our exercise, from the year 1996 to 2006.

The definite integral of a function is denoted by the integral symbol with lower and upper limits, which in our case are 0 and 10, representing the number of years since 1996. The process to find the total revenue is to integrate the revenue function, \(R(t)\), over that time period:
\[Total\_Revenue = \int_{0}^{10} R(t) \, dt\]
The calculations involve finding the antiderivative of the function and then applying the limits of integration to find the final value, which is often thought of as 'plugging in' the upper limit, subtracting the 'plugged in' lower limit, and interpreting the result—in this scenario, as the total revenue generated by the casino over 10 years.

Understanding and applying the concept of the definite integral is crucial for students, especially those involved in fields with practical applications such as economics, engineering, and the physical sciences.
Exponential Decay Model
Many phenomena in the real world, like depreciation of assets, radioactive decay, and certain types of revenue streams, can be modelled using functions that depict exponential decay. An \textbf{exponential decay model} is a mathematical representation where quantities decrease at a rate proportional to their current value.

In our given revenue function:
\[R(t) = 1,260 - 1,030 e^{-0.1 t}\]the component featuring an exponential function, \(e^{-0.1t}\), indicates that a portion of the revenue decreases over time, reflecting an aspect of the business model such as decreasing interest in a particular service or product over years. The rate of decay here is represented by the coefficient \(0.1\) in the exponent, which can be interpreted as a 10% decay rate per year.

When integrating this function to calculate total revenue, observing the exponential decay component's impact on the result is essential. It shows how initial values can be significantly higher, with the declining exponential function 'eating away' at that initial amount as time goes on. This lays the foundation for understanding more complex models in economics and finance, where such decay factors heavily into forecasting and evaluating current strategies.
Applied Calculus in Economics
Calculus, with its concepts of differentiation and integration, is not just confined to mathematics; it is a powerful tool in economics as well. \textbf{Applied calculus in economics} allows economists and business analysts to model and analyze dynamic changes in cost, revenue, and other economic indicators over time.

The use of integration to sum up revenues, costs, or investments over time, as illustrated in the casino revenue example, is one of many applications. By integrating the revenue function \(R(t)\), economists can approximate the total money made throughout a certain period, which in turn helps in setting future business goals, budgeting, and strategic planning.

Applied calculus serves as a bridge between theory and real-world applications, helping to make more informed and effective decisions in the management of resources. Economic models that incorporate calculus are integral to understanding market dynamics, optimizing resource allocation, and predicting how different variables will impact overall economic activity—a crucial consideration in every business and policy decision-making process.

Students of economics must thus recognize not just the methodology of integration but also how its results can interpret and influence real economic patterns and outcomes.

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Most popular questions from this chapter

According to the special theory of relativity, the apparent mass of an object depends on its velocity according to the formula $$ m=\frac{m_{0}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{1 / 2}} $$ where \(v\) is its velocity, \(m_{0}\) is the "rest mass" of the object (that is, its mass when \(v=0\) ), and \(c\) is the velocity of light: approximately \(3 \times 10^{8}\) meters per second. a. Show that, if \(p=m v\) is the momentum, $$ \frac{d p}{d v}=\frac{m_{0}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2}} $$ b. Use the integral formula for \(W\) in the preceding exercise, together with the result in part (a) to show that the work required to accelerate an object from a velocity of \(v_{0}\) to \(v_{1}\) is given by $$ W=\frac{m_{0} c^{2}}{\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}-\frac{m_{0} c^{2}}{\sqrt{1-\frac{v_{0}^{2}}{c^{2}}}} . $$ We call the quantity \(\frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\) the total relativistic energy of an object moving at velocity \(v\). Thus, the work to accelerate an object from one velocity to another is given by the change in its total relativistic energy. c. Deduce (as Albert Einstein did) that the total relativistic energy \(E\) of a body at rest with rest mass \(m\) is given by the famous equation $$ E=m c^{2} $$.

Evaluate the integrals. $$ \int_{-1}^{1} e^{2 x-1} d x $$

a. Show that the logistic function \(f(x)=\frac{N}{1+A b^{-x}}\) can be written in the form $$ f(x)=\frac{N b^{x}}{A+b^{x}} $$ HINT [See the note after Example 7 in Section 13.2.] b. Use the result of part (a) and a suitable substitution to show that $$ \int \frac{N}{1+A b^{-x}} d x=\frac{N \ln \left(A+b^{x}\right)}{\ln b}+C $$ c. The rate of graduation of private high school students in the United States for the period 1994-2008 was approximately $$ r(t)=220+\frac{110}{1+3.8(1.27)^{-t}} \text { thousand students per year } $$ \(t\) years since \(1994 .^{51}\) Use the result of part (b) to estimate the total number of private high school graduates over the period 2000-2008.

Evaluate the integrals. $$ \int_{0}^{1}\left(12 x^{5}+5 x^{4}-6 x^{2}+4\right) d x $$

(Compare Exercise 80 in Section 13.2.) The number of research articles in the prominent journal Physical Review written by researchers in the United States can be approximated by \(U(t)=\frac{4.6 e^{0.6 t}}{0.4+e^{0.6 t}}\) thousand articles per year \((t \geq 0)\) where \(t\) is time in years \((t=0\) represents 1983\() .^{50}\) Use a definite integral to estimate the total number of articles written by researchers in the United States from 1983 to \(2003 .\) HINT [See Example 7 in Section 13.2.]

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