Problem 24 - 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 \(x=b\). $$f(x)=x^{2}, g(x)=\frac{1}{x^{2}}, a=1, b=3$$

Step-by-Step Solution

Answer

The functions $f(x)=x^2$ and $g(x)=\frac{1}{x^2}$ intersect at x = 1. The area of the region enclosed between these two functions and the vertical lines x=1 and x=3 is found by taking the difference of the integrals of the functions over this interval: \[Area = \int_{1}^{3} (x^2 - \frac{1}{x^2}) dx\] Evaluating the integral, we get an area of 8 square units.

Step 1: Graphing the functions

Let's start by graphing the functions f(x) = x^2 and g(x) = 1/x^2. We know that f(x) = x^2 is a simple parabola with a vertex at the origin, and g(x) = 1/x^2 is a hyperbola that approaches but never touches the x and y-axis.

Step 2: Finding the points of intersection

Now, we need to find the points where f(x) and g(x) intersect. We can do this by setting f(x) equal to g(x) and solving for x: \(x^2 = \frac{1}{x^2}\) \(x^4 = 1\) \[x = \pm 1\] Thus, the curves intersect at the points x = 1 and x = -1. Since we are given a = 1 and b = 3, we only need to consider the intersection point of x = 1.

Step 3: Identifying the region enclosed by the graphs and the vertical lines

We will now determine the region enclosed between the vertical lines x = 1 and x = 3 and the graphs of f(x) and g(x). The region is enclosed by the parabola from x = 1 to x = 3 on the above side, and the hyperbola from x = 1 to x = 3 on the below side.

Step 4: Computing the area using integration

To find the area of the enclosed region, we can subtract the integral of g(x) from the integral of f(x) between x = 1 and x = 3. To do this, we set up the following integral: \[Area = \int_{1}^{3} (f(x) - g(x)) dx\] \[Area = \int_{1}^{3} (x^2 - \frac{1}{x^2}) dx\] Now, we can integrate the functions: \[Area = \left[\frac{x^3}{3} + \frac{1}{x}\right]_1^3\] And then evaluate the definite integral: \[Area = \left(\frac{3^3}{3} + \frac{1}{3}\right) - \left(\frac{1^3}{3} + 1\right)\] \[Area = \left(\frac{27}{3} + \frac{1}{3}\right) - \left(\frac{1}{3} + 1\right)\] \[Area = (9 + \frac{1}{3}) - (\frac{1}{3} + 1)\] \[Area = 8\] Thus, the area of the region enclosed between the vertical lines x = 1 and x = 3 and the graphs of f(x) and g(x) is 8 square units.

Key Concepts

Definite Integrals

In calculus, the definite integral of a function is a way to find the area under a curve between two points on the x-axis. It not only helps in calculating areas, but also in finding accumulated quantities. The definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated.

The result of a definite integral is a number, representing the total area. In solving the given exercise, we calculated the area between two curves by subtracting one definite integral from another. This subtraction gives the area between the curves within the given boundaries, which were the vertical lines at \( x = 1 \) and \( x = 3 \).

• To solve this, subtract the integral of the lower function \( g(x) = \frac{1}{x^2} \) from the integral of the upper function \( f(x) = x^2 \) in the region of interest.
• The solution calculates: \( \int_{1}^{3} (x^2 - \frac{1}{x^2})\, dx \).
• This leads to finding \( \frac{x^3}{3} + \frac{1}{x} \), evaluated from 1 to 3.

Using these definite integrals makes it easier to understand and compute complex areas involving curves, showing how powerful calculus can be.

Functions Graphing

Graphing functions is a foundational skill in calculus that helps visualize the behavior of mathematical equations. The function \( f(x) = x^2 \) represents a parabola opening upwards. Its symmetric curve touches the origin, and as the value of \( x \) increases, the square grows rapidly. On the other hand, \( g(x) = \frac{1}{x^2} \) is a hyperbola, approaching the x-axis and y-axis asymptotically. This means that as \( x \) moves away from zero, \( g(x) \) gets closer to the axes, but never actually touches them.

When graphing, it's important to note where these functions intersect since this is crucial to defining areas between them. In this example, setting \( f(x) = g(x) \) lets us solve for points where the curves meet. Solving \( x^2 = \frac{1}{x^2} \), we find they intersect at \( x = 1 \) and \( x = -1 \).

• The function \( f(x) \), due to its parabolic shape, is above \( g(x) \) between \( x = 1 \text{ and } x = 3 \), thus defining the region's boundary.
• The hyperbola, \( g(x) \), forms the below boundary up to \( x = 3 \).

These graphs help in visualizing how these functions behave and interact, making it easier to solve related calculus problems.

Area Between Curves

Finding the area between curves is a common problem in calculus. It requires utilizing definite integrals to determine the space between two curves across a given span of the x-axis. For the provided exercise, this concept facilitated calculating the area enclosed by \( f(x) = x^2 \) and \( g(x) = \frac{1}{x^2} \) between \( x = 1 \) and \( x = 3 \).

To find the area between curves:

• Identify the upper and lower functions within the given interval. For \( x = 1 \) to \( x = 3 \), \( f(x) = x^2 \) is the upper function, and \( g(x) = \frac{1}{x^2} \) is the lower.
• Establish the integral setup: \( \int_{1}^{3} (f(x) - g(x)) \, dx \).
• Calculate the integral for both functions separately, then subtract according to the order (upper minus lower).

The calculated result, 8 square units, demonstrates the area of the region enclosed, effectively using integration to understand interactions between curves. This technique is valuable for solving a variety of problems where different curves enclose areas.