Problem 15 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=x^{2}-4 x+3, \quad y=-x^{2}+2 x+3 $$

Step-by-Step Solution

Answer

The given functions intersect at the points (0,3) and (3,0). The function \(y = -x^2 + 2x + 3\) lies above the function \(y = x^2 - 4x + 3\) in the enclosed region. The area of the region is given by the integral: \(A = \int_0^3 (-2x^2 + 6x) \, dx\) Computing the integral, we find that the area enclosed by the two functions is 9 square units.

Step 1: Find the Points of Intersection

To find the points of intersection between the two functions, we must set them equal to each other: \(x^2 - 4x + 3 = -x^2 + 2x + 3\) Now, add \(x^2\) to both sides and subtract 2x from both sides to rewrite the equation as: \(2x^2 - 6x = 0\) To solve for x, first factor out the common factor of 2: \(2(x^2 - 3x) = 0\) Now, factor the quadratic expression within the parentheses: \(2(x)(x - 3) = 0\) Thus, the intersection points occur at x = 0 and x = 3. We can find the corresponding y-values by plugging these x-values into any of the functions. Let's use the first function: When x = 0, \(y = (0)^2 - 4(0) + 3 = 3\) When x = 3, \(y = (3)^2 - 4(3) + 3 = 0\) So, the points of intersection are at (0,3) and (3,0).

Step 2: Determine Which Function Is Above the Other

To find the area enclosed by these functions, we need to know which function is upper and which one is lower in this region. Find a point between the intersection points x = 0 and x = 3; for example, the midpoint x = 1.5. Then, compare the function values at x = 1.5: \( y_1 = 1.5^2 - 4(1.5) + 3 \approx -0.25 \) \( y_2 = -1.5^2 + 2(1.5) + 3 \approx 0.25 \) Since \(y_2 > y_1\), the second function (\(-x^2 + 2x + 3\)) is the upper function and the first function (\(x^2 - 4x + 3\)) is the lower function between the intersection points.

Step 3: Integrate the Difference of Functions

To find the area, we need to integrate the difference between the upper and lower functions from the first intersection point to the second intersection point: \(A = \int_a^b [(-x^2 + 2x + 3) - (x^2 - 4x + 3)] \, dx\) Simplify the expression inside the integral: \(A = \int_a^b (-2x^2 + 6x) \, dx\) Since we found the intersection points in Step 1, we know that \(a = 0\) and \(b = 3\). Plug these into the integral: \(A = \int_0^3 (-2x^2 + 6x) \, dx\)

Step 4: Evaluate the Integral

Compute the integral and find the area: \(A = \left[-\frac{2}{3}x^3 + 3x^2\right]_0^3\) \(A = -\frac{2}{3}(3)^3 + 3(3)^2 - (-\frac{2}{3}(0)^3 + 3(0)^2)\) \(A = -18 + 27\) \(A = 9\) Therefore, the area enclosed by the two functions is 9 square units.

Key Concepts

Integration

When we talk about integration in the context of mathematics, we're exploring a fundamental concept integral to calculus. Integration is the process used to find the area under a curve defined by a mathematical function. This process directly relates to the concept of finding the area between curves when those curves are defined by different functions.

In the exercise given, integration becomes a tool that allows us to calculate the area between two parabolic functions. It's like taking an oddly shaped piece of land and working out how big it is by filling it up with an infinite number of infinitely thin layers of paint, stacked one on top of the other. By integrating the difference between two functions over a certain interval, we determine the 'volume' of paint needed, which actually corresponds to the area between the curves.

Intersection Points of Functions

Anytime you're working with graphs of functions on a coordinate plane, there are points where the graphs intersect or cross each other. These are aptly called the intersection points of functions. The location of these points is found by setting the equations of the functions equal to each other and solving for the variable.

Determining the intersection points is crucial when calculating the area between curves because these points typically serve as the boundaries for the interval of integration. As shown in the exercise, after finding the intersection points, we know precisely where to start and stop our integration process as we attempt to find the area trapped between the curves.

Graph of Functions

The graph of a function is a visual representation of all the points that satisfy the function. Imagine you're plotting a road trip on a map, marking all the places where certain conditions are met—this is similar to graphing a function, but instead of cities and roads, we use coordinates and curves or lines.

Understanding the graph of the functions in question is instrumental for visualizing the region whose area we wish to compute. In the provided exercise, graphing both parabolic functions reveals where they behave differently and their relative positions to each other in the region of interest. Knowing which function lies above the other informs us about the nature of the area we intend to find.

Definite Integral

A definite integral is a type of integration that calculates the net area under a curve between two specific points, those points being the 'limits' of the integral. It's like specifying a distinct start and end point on a hiking trail and wanting to know the exact length of the path between them.

In practice, computing a definite integral involves finding the antiderivative of the function (essentially the 'reverse' of differentiation), then subtracting the value of this antiderivative at the lower limit from the value at the upper limit. This calculation method provides the exact area enclosed between the curve represented by the function and the horizontal axis, as we did with the exercise between the two parabolic curves.