Problem 1 - Calculus of a Single Variable

Question

Writing a Definite Integral In Exercises \(1-6,\) set up the definite integral that gives the area of the region. $$ \begin{array}{l}{y_{1}=x^{2}-6 x} \\ {y_{2}=0}\end{array} $$

Step-by-Step Solution

Answer

\(\int_{0}^{6} (x^2 - 6x) dx\).

Step 1: Identify the Points of Intersection

Set \(y_1\) equal to \(y_2\) and solve for \(x\). \n\n Given: \n\n \(y_1 = x^2 - 6x\) \n \(y_2 = 0\) \n\n Solve for \(x\): \n\n \(x^2 - 6x = 0\) \n \(x(x-6) = 0\)\n \(x = 0, 6\) \\n Thus, the points of intersection are at \(x = 0\) and \(x = 6\).

Step 2: Integrate the Difference of Functions

The area of the region defined by the two functions over the interval \([0,6]\) is given by the integral of the absolute difference of \(y_1\) and \(y_2\) over this interval. Thus we have: \n\n \(\int_{0}^{6} |x^2 - 6x - 0| dx\)

Step 3: Simplify the Integral

Since \(y_1\) is greater than \(y_2\) over the entire interval, we can remove the absolute value signs. Moreover, since \(y_2\) equals to zero, we can simplify the integral as follows:\n \(\int_{0}^{6} (x^2 - 6x) dx\).

Key Concepts

Area of a Region

To find the area of a region between two curves, we use the concept of definite integrals. This area can be visualized as the space bounded by the curves and the x-axis. The definite integral provides a way to calculate this space by integrating the function that represents the upper curve minus the function that represents the lower curve over a certain interval.
For instance, when dealing with the functions given, where one curve is defined by the equation \(y_1 = x^2 - 6x\) and the other is \(y_2 = 0\), we set up the definite integral of the difference of these functions over their intersection points. This difference gives the height of the slices of the region, and by summing them up over the interval, we find the area.

Points of Intersection

Determining the points of intersection between the two curves is crucial in setting up the limits for the definite integral. Points of intersection occur where the curves touch or cross each other. This happens when the functions are equal.
In our specific problem, solving the equation \(x^2 - 6x = 0\), gives us the intersection points \(x = 0\) and \(x = 6\). These points indicate the bounds of integration and are essential in defining the correct interval over which we will calculate the area between the curves.

Integration Techniques

To successfully integrate and find the area between curves, it's important to understand different integration techniques. These methods can include basic polynomial integration, substitution, and sometimes parts for more complex functions. In our exercise, the function \(x^2 - 6x\) is a simple polynomial, encouraging direct integration.
The integration is straightforward:

• The integral of \(x^2\) is \(\frac{x^3}{3}\).
• The integral of \(-6x\) is \(-3x^2\).

By applying these rules, we sum the areas of the narrow rectangles from \(x = 0\) to \(x = 6\) to find the total area.

Difference of Functions

The area between the curves is found using the difference of functions: \(y_1(x) - y_2(x)\). This difference represents the vertical distance between the curves at each point along the x-axis. Since \(y_2 = 0\), our focus is on \(y_1 - y_2\), or simply, \(y_1\).
The function \(y_1 = x^2 - 6x\) is subtracted from zero (which is essentially just the function itself in this case).
This gives the net height of the region at each point. Integrating this height over the interval \([0, 6]\) allows us to determine the entire area between the curves, highlighting the importance of evaluating and simplifying the difference when multiple functions are involved.