Problem 65 - Calculus of a Single Variable

Question

Area Between Curves The graphs of \(y=1-x^{2}\) and \(y=x^{4}-2 x^{2}+1\) intersect at three points. However, the area between the curves can be found by a single integral. Explain why this is so, and write an integral for this area.

Step-by-Step Solution

Answer

The integral which provides the area between the curves is \(A = \int_{-1}^{1} (1-x^{2} - (x^{4}-2 x^{2}+1)) dx\).

Step 1: Find the Points of Intersection

Solve \(1-x^{2} = x^{4}-2 x^{2}+1\) to get the intersection points. Since the equation simplifies to \(x^{4}-x^{2} = 0\), factoring provides \(x^{2}(x^{2}-1) = 0\). Hence, the solutions are \(x = -1, 0, 1\), which represent the points of intersection between the two curves.

Step 2: Determine the Curve Order

On the interval \(-1 \leq x \leq 1\), the curve \(y=1-x^{2}\) is above \(y=x^{4}-2 x^{2}+1\). To ensure the area between the two curves is positive, subtract the lower curve from the upper curve.

Step 3: Compute the Area Between the Curves

To compute the area between the two curves, use the formula \(A = \int_{a}^{b} |f(x) - g(x)| dx\). Here, \(f(x) = 1-x^{2}\), \(g(x) = x^{4}-2 x^{2}+1\), and the points of intersection \(a = -1\) and \(b = 1\) provide the interval of integration. Hence, the area is \(A = \int_{-1}^{1} (1-x^{2} - (x^{4}-2 x^{2}+1)) dx\).

Key Concepts

Definite Integration

Definite integration is a fundamental concept in calculus that deals with calculating the area under a curve between two points on the x-axis. This technique is crucial when figuring out the space trapped between two intersecting curves, such as those given in our exercise.

In the context of the problem, the integral is set up with limits of integration that correspond to the x-values of the intersecting points, in this case, -1 and 1. These limits ensure that we are only considering the area in the region of interest. Definite integration answers the question, 'How much area is there under curve f(x) between x=a and x=b?'

To properly execute definite integration, one typically follows these steps:

• Determine the integral of the function, known as the antiderivative.
• Apply the limits of integration (the intersection points) to the antiderivative.
• Calculate the difference between the upper limit evaluation and the lower limit evaluation.

The significance of the absolute value in the integral expression \( A = \int_{a}^{b} |f(x) - g(x)| dx \) is to ensure that the area calculated is positive, even if the function values are negative within the interval. This concept is particularly useful in situations where we are interested in the space between curves rather than the total area under a single curve.

Intersection Points in Calculus

Intersection points in calculus are points where two or more graphs meet, sharing common x and y values. These points are significant because they often demarcate the boundaries of an area we wish to calculate.

Discovering the points of intersection involves setting the equations of the curves equal to each other and solving the resulting equation for the x-values, which indicate where the curves cross. Essentially, finding these points is the first step in understanding the relationship between the curves.

In our exercise example, by solving \( 1-x^{2} = x^{4}-2 x^{2}+1 \), we determined the intersection points are at x=-1, 0, and 1. These results divide the x-axis into segments within which we can integrate to find the area between the curves. It is important to note that while the graphs intersect at three points, the relevant section for the area calculations is between x=-1 and x=1, as the curves do not cross between these points. This allows for the calculation of the area within that continuous interval using a single integral.

Integrals in Calculus

Integrals in calculus are primarily used to find areas, volumes, central points, and many useful things. However, integrals can be broken down into two main categories: indefinite and definite integrals.

An indefinite integral, represented by the symbol \( \int f(x) dx \), is a general form of an antiderivative and includes a constant of integration. In contrast, a definite integral, which we focused on in this exercise, has specific limits and provides a numerical answer representing the exact area.

The importance of integrals in calculus extends beyond merely finding areas. They are also used in physics for calculating quantities such as work and energy, in economics to find consumer and producer surplus, and in probability to determine the likelihood of certain outcomes, among other applications. In our exercise, we use integrals to measure the two-dimensional space between curves, which is a straightforward yet powerful illustration of the practicality of integrals in mathematical analysis.