Problem 15 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The region \(A\) in the first quadrant is bounded by \(y=2 x, y=-3 x+10\), and \(y=-\frac{1}{9}\left(x^{2}-6 x\right) .\) It has corners at \((0,0),(2,4)\), and \((3,1) .\) Express the area of \(A\) is the sum or difference of definite integrals. You need not evaluate.

Step-by-Step Solution

Answer

The area \(A\) expressed as the difference of definite integrals is: \(\int_{0}^{2} (2x - - \frac{1}{9}(x^{2}-6x)) dx + \int_{2}^{3} (-3x + 10 - - \frac{1}{9}(x^{2}-6x)) dx\)

Step 1: Plot and Identify the Ranges

First, plot the functions \(y=2x\), \(y=-3x+10\), and \(y=-\frac{1}{9}(x^{2}-6x)\). Identify and verify the points of intersection as provided in the problem, that is (0,0), (2,4), and (3,1). The range for each integral is then determined by these intersecting points of the functions.

Step 2: Determine Which Function is on Top

In order to know which function to subtract from which, determine which function lies on top. From the graph, it is seen that in the range [0,2], \(y=2x\) lies above \(y=-\frac{1}{9}(x^{2}-6x)\) and in the range [2,3], \(y=-3x+10\) lies above \(y=-\frac{1}{9}(x^{2}-6x)\).

Step 3: Setup and Express the Integral

Express the area of A as the sum or difference of two distinct integrals, with one integral ranging from x=0 to x=2, and the other from x=2 to x=3. The integrals represent the area 'under' one curve and 'above' another curve hence subtract one from another. So, for A: \(\int_{0}^{2} (2x - - \frac{1}{9}(x^{2}-6x)) dx + \int_{2}^{3} (-3x + 10 - - \frac{1}{9}(x^{2}-6x)) dx\)

Key Concepts

Area Between Curves

Understanding how to find the area between curves using definite integrals is a foundational concept in calculus. Essentially, this process involves calculating the 'net' area between two graphs. To visualize this, imagine the space that is filled when one curve is placed above another on a coordinate plane. The calculations become a comparison of the 'heights' of the functions, which are just the values of the functions at any given point along the x-axis.

When these functions cross over each other, they form a bounded region. The area of this region can be found by taking the integral of the upper function and subtracting the integral of the lower function between the points where the curves intersect. This subtraction is crucial as it effectively removes the surplus area counted twice when two separate integrals are taken.

Functions and Intersection Points

Before setting up the integrals to calculate the area between curves, it's essential to identify the intersection points of the functions. These points mark the limits where the area of interest begins and ends. The exercise given takes the functions in their algebraic form and provides the intersection points to make things clearer.

Identifying these points is typically done by setting the functions equal to each other and solving for x. Once found, these points become valuable in determining the limits of integration. Understanding the roles that intersection points play in boundary setting is a critical analytical step in solving area problems in calculus.

Setting Up Integrals for Area

The setup of integrals is a step-by-step process that requires attention to detail. Knowing which function is on top and which is at the bottom, over the interval of integration, is vital. To set up integrals for the area, you subtract the lower function from the upper function and integrate over the interval defined by the intersection points.

As illustrated in the exercise, two distinct integrals are set up because the 'top' and 'bottom' functions switch between the intervals [0,2] and [2,3]. In this case, you are dealing with piecewise-defined areas, which are treated separately and then summed together to find the total area between the curves.

Limits of Integration

The limits of integration are the x-values that bound the area of interest. They correspond to the intersection points of the functions involved. These limits tell us where to start and stop the integration process. In cases where there are multiple regions of intersection, like the triangle and trapezoid in the given problem, each segment's area is calculated using its specific limits of integration.

In the example, the integration limits are from x = 0 to x = 2 for the first integral and from x = 2 to x = 3 for the second integral, which matches the ranges where the upper and lower functions change. This careful segmentation ensures an accurate calculation of the area between curves with complex intersections.