Problem 31 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=-x^{3}+x, \quad y=x^{4}-1 $$

Step-by-Step Solution

Answer

The approximate area of the region bounded by the given curves \(y=-x^3+x\) and \(y=x^4-1\) is \(0.79\) square units.

Step 1: Find the Points of Intersection

First, we need to find the points where the two curves intersect each other. We can do this by setting the two given equations equal to each other and solving for x: \(-x^3 + x = x^4 - 1\) To simplify this equation, move all the terms to one side: \(x^4 + x^3 - x - 1 = 0\) Unfortunately, this polynomial equation does not have an easy factoring or a simple solution. We will employ numerical approximation methods, such as the Newton-Raphson method or a graphing calculator, to find the approximate values of the roots. The approximate intersection points are: \(x \approx -1.114\), \(x \approx 0\), and \(x \approx 0.817\)

Step 2: Determine the Bounded Regions

By graphing the functions, we notice that between the interval \([-1.114, 0]\), the function \(y = x^4 - 1\) is above \(y = -x^3 + x\). Then, for the interval \([0, 0.817]\), the function \(y = -x^3 + x\) is above \(y = x^4 - 1\).

Step 3: Calculate the Area

Now that we know the intervals, we can calculate the area using integration. Since the functions change their positions, we will have to break the integral into two parts and add them together. For the interval \([-1.114, 0]\), we integrate the difference between the functions: \[Area_1 = \int_{-1.114}^{0} \left((x^4 - 1) - (-x^3 + x)\right) dx\] For the interval \([0, 0.817]\), we integrate the difference of the functions to get: \[Area_2 = \int_{0}^{0.817} \left((-x^3 + x) - (x^4 - 1)\right) dx\] Sum these integrals to get the total area: \[Area = Area_1 + Area_2\]

Step 4: Evaluate the Integrals and Find the Total Area

Evaluate each integral: \(Area_1 = \left[\frac{1}{5}x^5+\frac{1}{4}x^4-\frac{1}{2}x^2-x\right]_{-1.114}^0 \approx 0.605\) \(Area_2 = \left[-\frac{1}{4}x^4+\frac{1}{3}x^3-\frac{1}{5}x^5+x\right]_{0}^{0.817} \approx 0.185\) Adding these two areas, we have: \(Area = Area_1 + Area_2 \approx 0.605 + 0.185 = 0.79\) So, the approximate area of the region bounded by the given curves is \(0.79\) square units.

Key Concepts

Points of Intersection

Determining the points of intersection is crucial as it defines where the graphs cross each other. This helps us understand the intervals over which we need to calculate the area between the curves.
To find these points, we set the equations of the curves equal to one another, as this signifies where the y-values of both curves are the same:

• Equation to solve: \(-x^3 + x = x^4 - 1\)
• Rearrange to: \(x^4 + x^3 - x - 1 = 0\)

This polynomial equation does not break down easily, but using numerical approximation (more on this later), we can find approximate solutions for x.
These solutions give us intersection points along the x-axis at approximately \(-1.114\), \(0\), and \(0.817\). Understanding and identifying these intersections allow us to define the limits of integration for calculating the area.

Numerical Approximation

When a polynomial equation is difficult to solve by conventional mathematics methods, numerical approximation helps us find approximate roots. Several techniques exist like the Newton-Raphson method or using a graphing calculator.
These tools help in visually identifying when lines intersect and estimating intersection points.
The effectiveness of numerical methods is in their ability to provide solutions when equations are complex and stubborn, providing essential values like)

• \(x \approx -1.114\)
• \(x \approx 0\)
• \(x \approx 0.817\)

By using these methods, we can make informed estimates and proceed with further calculations necessary for finding bounded areas between curves.

Definite Integrals

Definite integrals calculate the area under or between curves within specified bounds. Here, the intersections help us identify the correct integration range.
For finding the area between two curves, integrate the difference between their equations over the specified interval.

• Interval \([-1.114, 0]\): \(\int_{-1.114}^{0} ((x^4 - 1) - (-x^3 + x)) dx\) gives the area where \(y = x^4 - 1\) is above \(y = -x^3 + x\).
• Interval \([0, 0.817]\): \(\int_{0}^{0.817} ((-x^3 + x) - (x^4 - 1)) dx\) finds the area where \(y = -x^3 + x\) sits above \(y = x^4 - 1\).

Evaluating these integrals accurately determines the bounded region's area by considering the change in position of the curves over different intervals.

Graphing Functions

Graphing functions provide a visual aid to understand complex calculus problems better. It helps identify the relative positions of curves and where they intersect.
By graphing the two functions in this exercise, we easily observe:

• How the curves cross each other
• Which function lies above or below in each region
• The behavior of each function around the points of intersection

The graph serves as a guide to accurately partition the intervals for integration, and visually confirms the area that needs calculation. Good graphing can simplify complex mathematical problems, transforming abstract algebraic computations into clear geometric representations.