Problem 4 - Essential Calculus

Question

Set up the definite integral that gives the area of the region. $$ \begin{array}{l} f(x)=(x-1)^{3} \\ g(x)=x-1 \end{array} $$

Step-by-Step Solution

Answer

The definite integral that gives the area of the region between the two curves \(f(x)=(x-1)^3\) and \(g(x)=x-1\) from \(x=0\) to \(x=1\) is \(\int_{0}^{1} ( (x-1)^{3} - (x-1) ) dx\).

Step 1: Finding Points of Intersection

Set f(x) equal to g(x) to find the points where the two graphs intersect. So, set \((x-1)^{3}\) equal to \(x-1\). Solve for \(x\). This gives \(x=1, 0\).

Step 2: Determining the Order of Functions

Between \(x=0\) and \(x=1\), check which function is above the other. For example, plug in \(x = 0.5\) in both f(x) and g(x). Here, f(x) gives \(0.5 - 1 = -0.5\), and g(x) gives \((0.5-1)^{3} = -0.125\). Since \(g(x) > f(x)\) for \(x = 0.5\), in this range, the curve \(f(x)=(x-1)^{3}\) is above \(g(x)=x-1\).

Step 3: Integral Setup for Area Calculation

Setup the definite integral for the area between the curves from \(x = 0\) to \(x = 1\), defined as the absolute difference of the two functions. So, the integral is \(\int_{0}^{1}\mid f(x)-g(x) \mid dx = \int_{0}^{1} ( (x-1)^{3} - (x-1) ) dx\).

Key Concepts

Area Between Curves

When we want to find the area between two curves, we look for the region where the two functions overlap vertically. Imagine drawing two lines, one from each curve, down to the x-axis. The space in between these lines is what we aim to measure.

To calculate this area, we need to:

• Identify where the functions intersect each other on the x-axis.
• Determine which function lies above the other over the interval of interest.
• Calculate the area by setting up a definite integral using the difference of these functions.

Remember, the integral gives us a precise measurement by accumulating very small slices of this area. The slices are tallied up from one point of intersection to the other, ensuring that we accurately measure the entire region between the curves.

Points of Intersection

Finding where two curves intersect is crucial for determining the boundaries of the area between them. In mathematical terms, points of intersection are where the outputs (y-values) of the functions are equal for specific x-values.

In our exercise, we equate \((x-1)^3\) to \(x-1\) and solve for \(x\). This gives us the points \(x = 1\) and \(x = 0\). These intersections serve as the limits for our integral.

Once you've found the points of intersection, they guide us in setting the limits of integration, which are the end points of the interval from which we'll calculate the area between the curves.

Integral Calculation

An integral is a mathematical tool we use for adding up infinitely small quantities, typically to find areas or volumes. When calculating the area between curves, we set up a definite integral using the difference between the functions.

Here’s how we set it up:

• Determine the interval over which to integrate, using the points of intersection: \(x = 0\) and \(x = 1\).
• Identify which function is on top and which is below over this interval by comparing a middle value, such as \(x = 0.5\).
• Calculate the integral as \( \int_{0}^{1} \left( (x-1)^3 - (x-1) \right) \, dx \).

The integral, when evaluated, gives the area between the curves. It considers the net effect, adding slices where the top function dominates, ultimately yielding the area value.

Function Comparison

When examining two functions to find the area in between, it’s important to know which one is higher or lower over a given interval. This determines how we set up our integral.

You can compare by stepping into the interval and:

• Picking a value between the intersection points, such as \(x = 0.5\).
• Substitute this x-value into both functions.
• With \(f(x) = (x-1)^3\) and \(g(x) = x-1\), find that at \(x = 0.5\), \(g(x)\) is actually higher than \(f(x)\).

This information tells us how to subtract one from the other in the integral, ensuring we use positive differences for area calculation. Essentially, function comparison helps keep our math organized and our area calculations correct.