Problem 35 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graph and find the area of the region completely enclosed by the graphs of the given functions \(f\) and \(g\). $$f(x)=x+2\( and \)g(x)=x^{2}-4$$

Step-by-Step Solution

Answer

The two functions \(f(x) = x + 2\) and \(g(x) = x^2 - 4\) intersect at points A(-2,0) and B(3,5). Upon sketching the graphs, the region enclosed by the graphs is found between these two points, with \(f(x)\) above \(g(x)\). By integrating the difference between the functions with respect to x, from the leftmost intersection point to the rightmost intersection point, we can determine the enclosed area. The area is calculated as \(\frac{69}{6} = 11.5\) square units.

Step 1: Identify the intersection points

To find the points where the graphs intersect, we need to find the points where the functions are equal, i.e., when: \[f(x) = g(x)\] \[x + 2 = x^2 - 4\] Now we need to solve for the x values of the intersection points. Rewrite the equation as: \[x^2 - x - 6 = 0\] This is a quadratic equation, which can be factored: \[(x - 3)(x + 2) = 0\] This gives us two intersection points: \[x - 3 = 0 \Rightarrow x=3 \quad \text{and} \quad x + 2 = 0 \Rightarrow x=-2\] Now we can find the corresponding y values by plugging the x values back into one of the original functions, e.g., \(f(x)\): \[f(3) = 3 + 2 = 5\] \[f(-2) = -2 + 2 = 0\] So, the intersection points are A(-2,0) and B(3,5).

Step 2: Sketch the graph and identify the enclosed region

Now that we have both functions and the intersection points, we can sketch the graphs of the functions and identify the enclosed region. The function \(f(x) = x + 2\) is a linear function with a positive slope, and \(g(x) = x^2 - 4\) is a parabolic function facing upwards. The region enclosed by the graphs is found between A and B, with \(f(x)\) above \(g(x)\).

Step 3: Find the enclosed area using integration

To find the area of the enclosed region, we need to integrate the difference between the functions with respect to x, from the leftmost intersection point (-2) to the rightmost intersection point (3): \[Area = \int_{-2}^{3} (f(x) - g(x)) \, dx\] Next, we will substitute the functions for \(f(x)\) and \(g(x)\) and solve the integral: \[Area = \int_{-2}^{3} (x + 2 - (x^2 - 4)) \, dx\] \[Area = \int_{-2}^{3} (-x^2 + x + 6) \, dx\] Integrate the function with respect to x: \[Area = \frac{-x^3}{3} + \frac{1}{2}x^2 + 6x \Big|_{-2}^{3}\] Now, evaluate the definite integral: \[Area = \left[\frac{-3^3}{3} + \frac{1}{2}(3)^2 + 6(3) \right] - \left[\frac{-(-2)^3}{3} + \frac{1}{2}(-2)^2 + 6(-2) \right]\] \[Area = \left[-27 + \frac{9}{2} + 18 \right] - \left[\frac{-8}{3} + 2 - 12 \right]\] \[Area = \left[-\frac{45}{2} + \frac{17}{3} \right]\] \[Area = \frac{-69}{6}\] Since the area must be positive, we take the absolute value: \[Area = \frac{69}{6} = 11.5\] The area of the region completely enclosed by the functions \(f(x) = x + 2\) and \(g(x) = x^2 - 4\) is 11.5 square units.

Key Concepts

Definite Integral

When we talk about finding the area between two curves, the definite integral is a powerful tool used in calculus. It helps us quantify the 'accumulation' of quantities, which, in the case of area calculations, translates to finding the sum of infinitesimally thin rectangles between curves over a certain interval.

The process involves integrating the difference between the two functions that define the curves within the limits of integration. These limits are usually the points where the two curves intersect, which provide the boundaries for the area in question. In essence, a definite integral takes the form \[\int_{a}^{b} (f(x) - g(x)) \, dx\] where \(a\) and \(b\) are the bounds of integration, and \(f(x)\) and \(g(x)\) represent the functions for which we seek the enclosed area. The result is a numerical value that represents the total area, accounting for the fact that areas below the x-axis are considered negative by convention—hence the use of absolute value to ensure we report the area's magnitude only.

Graph Sketching

Visualizing the relationship between functions on a graph is key in understanding the behavior of mathematical equations—this is where graph sketching comes into play. A graph provides a picture of how functions increase, decrease, and where they intersect, which is particularly useful when determining areas between curves.

For instance, in our example we sketched a straight line for the linear function \(f(x) = x + 2\) and a parabola for the quadratic function \(g(x) = x^2 - 4\). Through sketching, we identified the region enclosed by these functions, which allowed us to visualize the area we were set to calculate. Getting familiar with different function types, such as linear or quadratic, and their respective shapes on a graph is an essential skill for succeeding in finding areas between curves.

Intersection Points

The concept of finding intersection points is crucial when dealing with the graphs of two functions. These points represent the x-values where the functions have the exact same y-value, implying that the curves intersect at these coordinates.

By setting the two functions equal to each other, we can solve for x to find the intersection points. These points ∗(denote the limits of the area we wish to calculate∗. In our exercise, the intersection points were found by solving the quadratic equation \(x^2 - x - 6 = 0\), leading to the points A(-2,0) and B(3,5). Once we have these points, we can proceed to calculate the area between curves with certainty that we're capturing the entire region of interest.

Quadratic Functions

A quadratic function is a second-degree polynomial of the form \(f(x) = ax^2 + bx + c\). Its graph is a parabola that can open upwards or downwards, depending on the sign of the coefficient a.

In the example problem, \(g(x) = x^2 - 4\) is a quadratic function where the parabola opens upwards since the coefficient of \(x^2\) is positive. The vertex of this parabola is the point at which the curve changes direction, and for a standard quadratic equation, this vertex can be found at the point \(\frac{-b}{2a}, g(\frac{-b}{2a})\). Understanding how to sketch and analyze quadratic functions helps immensely when identifying the regions between curves and enables us to apply integration effectively to find areas between such functions.