Problem 46 - Calculus Early Transcendentals

Question

Use the most efficient strategy for computing the area of the following regions. The region bounded by \(y=x^{3}, y=-x^{3},\) and \(3 y-7 x-10=0\)

Step-by-Step Solution

Answer

Answer: The most efficient strategy is to follow these steps: 1. Sketch the bounding functions on a coordinate plane. 2. Find the points of intersection between the line and the cubic functions. 3. Use integration to compute the area of the region bounded by the functions, by integrating the difference between the cubic functions and the line over the interval [A, B], where A and B are the points of intersection.

Step 1: Sketch the bounding functions

First, we need to visualize the given region by sketching the functions y = x³, y = -x³, and 3y - 7x - 10 = 0 on a coordinate plane. Since y = x³ is a cubic function, its graph is a curve with increasing steepness as x increases. Similarly, y = -x³ is also a cubic function, with increasing steepness as x decreases. The line 3y - 7x - 10 = 0 is a straight line, which we can write as y = (7/3)x + 10/3.

Step 2: Find points of intersection

To find the points of intersection between the line and the cubic functions, we need to solve the following systems of equations: 1. y = x³ and y = (7/3)x + 10/3 2. y = -x³ and y = (7/3)x + 10/3 For the first system, substitute the value of y from the first equation into the second equation: x³ = (7/3)x + 10/3 x³ - (7/3)x - 10/3 = 0 For the second system, substitute the value of y from the first equation into the second equation: -x³ = (7/3)x + 10/3 - x³ - (7/3)x - 10/3 = 0 Now, solve these equations to find the points of intersection. Let the points of intersection be A and B.

Step 3: Compute the area

To compute the area of the region bounded by the functions, we need to integrate the difference between the cubic functions and the line over the interval [A, B]. Area = \(\int_{x_A}^{x_B} ((x^3 - (-x^3)) - ((7/3)x + 10/3)) dx\) = \(\int_{x_A}^{x_B} (2x^3 - (7/3)x - 10/3) dx\) Now, calculate the integral and evaluate it with the limits of integration found in Step 2. The final area value is the result of this evaluation.

Key Concepts

Area under curve

Finding the area under a curve is a fundamental concept in calculus. It involves determining the region that lies between the graph of a function and the horizontal axis over a certain interval. This concept is vital as it has practical applications, such as calculating physical quantities like distance, volume, and probability. In our specific exercise, we are interested in the area bounded by the graphs of the cubic functions and the line. To effectively calculate this, we need to focus on the sections where these curves interact. Simply integrating the difference between the curves will give us the desired area. Here's the essential procedure:

• Identify the functions that form the boundary of the area in question.
• Determine the points of intersection between these functions to set the limits of integration.
• Calculate the integral of the upper function minus the lower function over the interval defined by these points.

This integral will quantify the area under the curve, providing valuable insights into the nature of the functions involved.

Integration

Integration is the mathematical process used to calculate areas, among other things. It's the opposite of differentiation and allows us to "accumulate" quantities. When you integrate a function, you're essentially adding up an infinite number of infinitesimally small areas to find a total area. Consider our exercise where we needed to find the bounded area. We perform integration on the expression that represents the difference between the curves. Generally, this is symbolized as: \[ \int_{a}^{b} f(x) \, dx \]In our case, the function for integration was \( 2x^3 - \frac{7}{3}x - \frac{10}{3} \) evaluated over a certain interval [A, B]. This process entails several steps:

• Determine the integrand, which is the function you are integrating.
• Set the limits of integration, corresponding to where the curves intersect.
• Calculate the definite integral to find the area.

By mastering integration, we can solve problems involving curves and lines, gaining more insight into the behavior of different mathematical models.

Cubic functions

Cubic functions, such as \( y = x^3 \), form a critical part of this problem. These are polynomials of degree three, and their graphs are characterized by their curve shape, which can change direction. They have one or two turns, depending on the specific form of the function. When dealing with cubic functions:

• Recognize that their shape is symmetric about the origin when no linear or constant terms are involved.
• Understand that they can have up to three real roots, where the graph intersects the x-axis.
• Pay attention to their steep slopes at the ends, as they rapidly grow positive or negative.

In the context of our exercise, \( y = x^3 \) and \( y = -x^3 \) stylize symmetrical curves that open upwards and downwards, respectively. By examining these functions' interactions, such as intersections with lines or other curves, we can determine critical points that inform subsequent calculations of bounded areas.

Intersection points

Finding intersection points is crucial in identifying the limits for integration when determining bounded areas. Intersection points are where two or more curves meet. In calculus, these points help establish segments over which integration is performed.For the given regions, the intersection between various functions sets our limits for integration. Here's what you need to know:

• To find intersections, solve the system of equations representing the curves.
• For example, if you have \( y = x^3 \) and \( y = \frac{7}{3}x + \frac{10}{3} \), set them equal and solve for \( x \).
• This calculation reveals the critical \( x \)-values that bound the area of interest.

In the current problem, we're interested in such intersections because they define the integration range, allowing us to compute the area accurately. Mastering the technique of finding intersection points aids in tackling various challenges in calculus, ultimately simplifying complex function analysis.