Question

The area of the region bounded by the graphs of \(y=x^{3}\) and \(y=x\) cannot be found by the single integral \(\int_{-1}^{1}\left(x^{3}-x\right) d x\). Explain why this is so. Use symmetry to write a single integral that does represent the area.

Short answer

The given integral will not accurately represent the area as it does not consider the absolute difference between the functions, incorrectly yielding negative values for positive x. By recognizing that the functions are symmetric about the y-axis, the correct integral becomes: \(2\int_{0}^{1} \left(x-x^{3}\right) dx\)

Step-by-step solution

Understand the relationship between the functions

Plot the graphs of both functions, \(y=x\) and \(y=x^{3}\). You will observe that from -1 to 1, \(x^{3}\) is below \(x\) for positive x values and above for negative x values. Thus, \(x^{3} - x\) would yield negative values for positive x and positive values for negative x, which is inconsistent with the concept of area being non-negative.

Setting up the correct integral for calculating area between curves

The area between two curves is found by integrating the absolute difference of the functions. So, the area bounded by the curves from -1 to 1 would be given by: \(\int_{-1}^{1} |x^{3}-x| dx\).

Utilize Symmetry

Given that the functions \(y=x\) and \(y=x^{3}\) are symmetric about the y-axis, the area under the curve from -1 to 0 is the same as the area from 0 to 1. Thus, to simplify, you can find the area from 0 to 1, then multiply by two. Consequently, the single integral that represents the area becomes: \(2\int_{0}^{1} \left(x-x^{3}\right) dx\)

Key concepts

Integrating Absolute Value Functions

When working with integrating absolute value functions, it's important to understand that the absolute value ensures all results of the function within the absolute value signs are non-negative.

This is especially critical when determining areas between curves as negative values would suggest subtracting area, which contradicts the fundamental goal of measuring the extent of a region. In the given exercise, the integral of the difference between the functions y=x and y=x^3 over the interval from -1 to 1 must account for this absolute value.

The procedure involves breaking up the integral into parts where the integrand is positive and parts where it is negative. Here’s how you can proceed:

• First, identify the points of intersection of the two curves, which act as the limits of integration.
• Next, determine the intervals on which x^3 is greater than x, and vice versa.
• For each interval, integrate the absolute value of the difference between x and x^3.
• Finally, add the results to get the total area.

This approach will provide the correct measurement of the region's area.

Symmetry in Calculus

The concept of symmetry in calculus is a powerful tool to simplify calculations. Symmetry can apply to both functions and regions bounded by these functions.

When a function or a region is symmetric about an axis, similar calculations on either side of the axis yield the same result. This symmetry is evident in this exercise since the graph of the functions y=x and y=x^3 shows mirror symmetry across the y-axis.

Recognizing this, one can compute the area of one symmetric half and double the result to find the total area, instead of computing the area across an interval where the function values may change sign. Utilizing symmetry:

• Reduces the complexity of problems
• Lowers computation time
• Eliminates the need for integrating across intervals where the integrand changes sign

Always look for symmetry in problems involving areas between curves; it can greatly streamline your approach to finding solutions.

Calculus Step by Step Solution

A calculus step by step solution is an invaluable methodology for students to grasp complex calculus concepts and understand the procedures involved in solving problems. It involves breaking down a calculus problem into manageable parts and solving each part one at a time.

For the exercise at hand, the step by step solution efficiently guides through the process:

• Understanding the relationship between the functions graphically
• Setting up the correct integral with consideration of absolute values
• Applying the concept of symmetry to simplify the computation

By following these enumerated steps, students can tackle each aspect of the problem methodically, which reinforces learning and ensures no detail is overlooked.

Whenever you face a calculus problem, especially one involving areas or integrals, try to dissect it into a series of logical steps. This habit not only makes complex problems manageable but also enhances your analytical skills and deepens your understanding of calculus principles.