Problem 3 - Essential Calculus

Question

Set up the definite integral that gives the area of the region. $$ \begin{array}{l} f(x)=3\left(x^{3}-x\right) \\ g(x)=0 \end{array} $$

Step-by-Step Solution

Answer

The definite integral that gives the area of the region is \[ \int_{-1}^{0}(3(x^3-x)-0) dx + \int_{0}^{1} (0 - 3(x^3-x)) dx \]

Step 1: Find the intersection points

Set the two equations equal to each other to find the points of intersection. Here, since g(x) = 0, simply set f(x) = 0, where \(3(x^3-x) = 0\). Solving this equation, x can be 0, 1, and -1

Step 2: Set up the integral

Now set up the definite integral with the right minus the left function as the integrand, and the intersection points as the limits of integration. In this case, since the function g(x) = 0 represents the x-axis and is less than f(x), the integral becomes \[\int_{-1}^{0} (3(x^3-x)-0) dx + \int_{0}^{1} (0 - 3(x^3-x)) dx \]. The integrals are separated because the function changes the side at \(x=0\).

Key Concepts

Integral Setup

The process of setting up an integral to calculate the area between two curves is crucial to solving many calculus problems. An integral is essentially a mathematical tool that allows us to sum up infinitesimally small quantities, which in the case of area calculations, are the tiny slices of area under a curve. When the equations of the curves are provided, as in the exercise with f(x) = 3(x^3 - x) and g(x) = 0, the task at hand is to integrate the difference between these two functions.

To set up the correct integral, we must understand which function is on top (greater value) and which one is below (lesser value) over the interval we're interested in. For the problem given, since g(x) represents the x-axis, it is always lower or equal to f(x) between the intersection points. The integrand – the function we're integrating – is therefore the difference between these two functions, f(x) - g(x), in the areas where f(x) is above the x-axis, and g(x) - f(x) where f(x) is below the x-axis.

Intersection Points

To determine the range over which the area needs to be calculated, we must find the points of intersection between the curves. These intersection points become the limits in the definite integral. In our exercise, we are asked to calculate the area between the curve f(x) = 3(x^3 - x) and the x-axis, g(x) = 0.

To find out where these two graphs intersect, we equate f(x) to g(x). This results in the equation 3(x^3 - x) = 0. Solving this cubic equation, we find the x-values where the curve meets the x-axis. In this case, the intersections occur at x = -1, x = 0, and x = 1. These points are essential because they split the area into two parts, with the curve above the x-axis from x = -1 to x = 0 and below from x = 0 to x = 1.

Limits of Integration

Once we've identified the intersection points, we're ready to establish the limits of integration. These are the values that we will use as the start and stop points on the x-axis for our integral. They are crucial because they delineate the exact section of the graph over which we want to calculate the area.

In the case of our exercise, the limits of integration are x = -1, x = 0, and x = 1. We use these limits to break up our integral into two parts since the integral changes sign when crossing over the x-axis at x = 0. Therefore, we calculate the integral from x = -1 to x = 0, then from x = 0 to x = 1. This allows us to accurately compute the total area by taking the absolute value of each integral, ensuring a positive result that corresponds to the actual physical area. The split integral hence reflects the change in relationship between the functions across the domain of interest.