Problem 25 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graphs of the functions \(f\) and \(g\) and find the area of the region enclosed by these graphs and the vertical lines \(x=a\) and \(x=b\). $$f(x)=e^{x}, g(x)=\frac{1}{x} ; a=1, b=2$$

Step-by-Step Solution

Answer

To find the area enclosed by the curves \(f(x)=e^x\) and \(g(x)=\frac{1}{x}\) and the vertical lines \(x=a=1\) and \(x=b=2\), we first find the intersection point \(c\) between the two functions. Then, we can find the area \(A_1\) between the two functions from \(x=a\) to \(x=c\), and the area \(A_2\) between the two functions from \(x=c\) to \(x=b\). Finally, we sum these two areas to find the total enclosed area: \(A = A_1 + A_2 = (e^c - e) + (\ln(2) - e^2)\).

Step 1: Sketch the graphs of \(f(x)\) and \(g(x)\)

To begin, let's sketch the graphs of \(f(x) = e^x\) and \(g(x) = \frac{1}{x}\). Recall that: - The graph of \(f(x) = e^x\) is an exponential function that is always positive and increasing. - The graph of \(g(x) = \frac{1}{x}\) is a hyperbolic function with a vertical asymptote at \(x = 0\) and a horizontal asymptote at \(y = 0\). By plotting these two functions, we will be able to visualize the area enclosed by the curves and the vertical lines.

Step 2: Identify the points of intersection

To find the points where the graphs of \(f(x)\) and \(g(x)\) intersect, we need to solve the equation \(f(x) = g(x)\) for \(x\), which gives us: $$e^x = \frac{1}{x}$$ Solving this equation analytically is not trivial. However, since we are only interested in the intersection points between \(a = 1\) and \(b = 2\), we can observe from the graph that there is one intersection point within this interval. After plotting the graphs, we can use a numerical method, like the bisection method or Newton's method, to approximate the intersection point, denoted as \(c\).

Step 3: Calculate the area enclosed between the functions

To find the area of the region enclosed by the two functions and the vertical lines \(x = a\) and \(x = b\), we will use integration. First, we calculate the area between the two functions from \(x = a\) to \(x = c\), and then the area between the two functions from \(x = c\) to \(x = b\). Since \(f(x) > g(x)\) in the interval \([a, c]\), we will find the area by integrating the difference of the two functions: $$A_1 = \int_{a}^{c} [f(x) - g(x)] dx = \int_{1}^{c} [e^x - \frac{1}{x}] dx$$ Similarly, we can find the area of the region enclosed by the two functions from \(c\) to \(b\): $$A_2 = \int_{c}^{b} [g(x) - f(x)] dx = \int_{c}^{2} [\frac{1}{x} - e^x] dx$$ Finally, the total area enclosed by the functions and the vertical lines is the sum of \(A_1\) and \(A_2\): $$A = A_1 + A_2 = \int_{1}^{c} [e^x - \frac{1}{x}] dx + \int_{c}^{2} [\frac{1}{x} - e^x] dx$$ $$A = \left[ e^x + \ln(x) \right]_1^c + \left[ \ln(x) - e^x \right]_c^2$$ $$A = (e^c + \ln(c) - e + \ln(1)) + (\ln(2) - e^2 - \ln(c) + e^c)$$ $$A = (e^c - e) + (\ln(2) - e^2)$$ After estimating the intersection point \(c\), we can plug its value into the formula above to find the area of the enclosed region.

Key Concepts

Exponential Function

Exponential functions are mathematical functions of the form \(f(x) = a^{x}\), where \(a\) is a positive constant base, and \(x\) is the exponent. A unique characteristic of exponential functions is their rapid rate of increase. As \(x\) increases, \(e^x\) also increases. This rapid growth rate leads to curves that climb upwards continuously as we move to the right along the horizontal axis.

• The base of the natural exponential function is Euler's number, \(e\), which is approximately 2.718. This value gives rise to the specific exponential function \(f(x) = e^x\).
• Exponential functions, like \(f(x) = e^x\), are always positive because \(e\) raised to any power cannot yield a negative result.
• They have notable applications in growth models, compound interest, and calculus itself.

Visualizing an exponential function often helps to understand its behavior over an interval. In our exercise, the graph of \(f(x) = e^x\) helps us picture part of the area enclosed by the given curves and vertical lines between \(x = 1\) and \(x = 2\).

The exponential function's graph doesn't have any x-intercepts because the function never touches the horizontal axis. Instead, it continues rising sharply as \(x\) becomes larger.

Hyperbolic Function

Hyperbolic functions are analogues of trigonometric functions but based on hyperbolas rather than circles. Unlike trigonometric functions, they do not oscillate; instead, they present unique characteristics.

The simplest example of a hyperbolic function is \(g(x) = \frac{1}{x}\), often referred to as the reciprocal function. This function is important in analyzing real-world scenarios involving inverse relationships.

• The graph of \(g(x) = \frac{1}{x}\) has vertical asymptotes, usually occurring at \(x = 0\). This is because the function is undefined there, leading to a break in the graph.
• Horizontal asymptotes appear at \(y = 0\), showing the values \(g(x)\) approaches as \(x\) becomes very large or very different from zero.
• Hyperbolic functions have applications in geometry and are useful in situations where inverse proportionality is present.

In the exercise, the hyperbolic function \(g(x) = \frac{1}{x}\) is part of the bounded region formed with \(f(x)\) between \(x = 1\) and \(x = 2\). This helps with determining specific areas formed by the two intersecting graphs.

Definite Integral

A definite integral is a fundamental concept in calculus used to compute the area under curves within a specified interval on the x-axis, typically noted as \(\int_{a}^{b} f(x) \, dx\). This allows us to calculate areas enclosed by curves and lines, like the problem at hand where \(f(x)\) and \(g(x)\) enclose a region between \(x = 1\) and \(x = 2\).

The process of finding a definite integral involves:

• Identifying the function representing the curve, such as \(e^x\) and \(\frac{1}{x}\) in our problem.
• Establishing integration limits. Here, it is between the vertical lines \(x = 1\) and \(x = 2\).
• Calculating the integral's value using the Fundamental Theorem of Calculus, which assures that the definite integral can be evaluated using antiderivatives.

This yields the total area, calculated by integrating the absolute difference of \(f(x)\) and \(g(x)\).

Understanding definite integrals is key in contexts such as physics for determining work done by a force, and in probability for finding total probabilities. In the equation provided, the intersection point \(c\) requires numerical methods to be determined accurately, which shows the dynamic interaction of an exponential and hyperbolic function on an interval.