Problem 19 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graphs of the functions \(f\) and \(g\) and find the area of the region enclosed by these graphs and the vertical lines \(x=a\) and \(x=b\). $$f(x)=-x^{2}+2 x+3, g(x)=-x+3 ; a=0, b=2$$

Step-by-Step Solution

Answer

The graphs of \(f(x) = -x^2 + 2x + 3\) and \(g(x) = -x + 3\) intersect at \(x=0\). The area of the region enclosed by these functions and vertical lines \(x=0\) and \(x=2\) is found by integrating the difference between the functions from 0 to 2: \[Area = \int_0^2 (-x^2 + 3x) dx\]. The enclosed area is \(\frac{16}{3}\).

Step 1: Sketch the graphs of f and g

Begin by sketching the graphs of both functions: 1. The quadratic function \(f(x) = -x^2 + 2x + 3\) has a negative leading coefficient (-1) and therefore has a U-shaped graph that opens downwards. The vertex of the parabola is the maximum point and can be found using the formula \(x = \frac{-b}{2a}\), where \(a = -1\) and \(b = 2\). Here, the vertex is at \(x = \frac{-2}{2(-1)} = 1\). 2. The linear function \(g(x) = -x + 3\) has a negative slope (-1) and a positive y-intercept (3), indicating that the line decreases as x increases, with a starting point of (0,3).

Step 2: Find the points of intersection

Set f(x) and g(x) equal to one another to find the points of intersection: \(-x^2 + 2x + 3 = -x + 3\) Add \(x^2\) to both sides and subtract in order to get: \( x^2 + 3x = 0\) Factor out \(x\): \(x(x + 3) = 0\) Set each factor equal to zero and solve for x: \(x = 0\) \(x + 3 = 0 => x = -3\) However, since we're only considering the area between x=a (0) and x=b (2), we will only use the intersection at x=0.

Step 3: Find the area of the enclosed region

In order to find the area of the region enclosed by the functions f and g and the vertical lines x=a and x=b, integrate the difference between f(x) and g(x) from a to b: \[Area = \int_a^b (f(x) - g(x)) dx\] Our bounds are a=0 and b=2, and our functions are \(f(x) = -x^2 + 2x + 3\) and \(g(x) = -x + 3\). Plug these into the equation: \[Area = \int_0^2 ((-x^2 + 2x + 3) - (-x + 3)) dx\] Simplify the integrand: \[Area = \int_0^2 (-x^2 + 2x + 3 + x - 3) dx\] \[Area = \int_0^2 (-x^2 + 3x) dx\] Integrate the result: \[Area = \left[-\frac{1}{3}x^3 + \frac{3}{2}x^2 \right]_0^2\] Evaluate the integral at the limits: \[Area = \left(-\frac{1}{3}(2)^3 + \frac{3}{2}(2)^2\right) - \left(-\frac{1}{3}(0)^3 + \frac{3}{2}(0)^2\right)\] \[Area = \left(-\frac{8}{3} + 6\right) - (0)\] \[Area = -\frac{2}{3} + 6 = \boxed{\frac{16}{3}}\] The area of the region enclosed by the graphs of functions f and g and the vertical lines x=0 and x=2 is \(\frac{16}{3}\).

Key Concepts

Graphing Functions

When it comes to visualizing the behavior of functions, graphing is a key tool in calculus. To plot a function like \( f(x) = -x^2 + 2x + 3 \), we follow a systematic approach:

• Identify the type of function: Here, \( f(x) \) is a quadratic function because it takes the form of \( ax^2 + bx + c \).
• Determine the direction of the graph: The coefficient of \( x^2 \) is -1, so the parabola opens downward.
• Find the vertex: The formula \( x = \frac{-b}{2a} \) helps us find the x-coordinate of the vertex. In this case, it's \( x = 1 \).
• Plot points for clarity: Besides the vertex, you can plot other points by selecting x-values and calculating corresponding y-values.

Similarly, linear functions like \( g(x) = -x + 3 \) are easier to graph:

• Slope and intercept form: \( g(x) \) is in the form of \( y = mx + c \), where \( m = -1 \) and \( c = 3 \).
• Direction of the line: The negative slope indicates a decreasing line.
• Intercept plotting: Start at the y-intercept (0,3) and decrease one unit on y for every unit increase on x.

Understanding these basic rules simplifies the graphing process and makes further analysis clearer.

Integral Calculus

Integral calculus is a fundamental tool used to find areas, among other things. Here, we're interested in calculating the area between two curves. We do this through integration:

• Set up the integral: The formula \[ Area = \int_a^b (f(x) - g(x)) \, dx \] gives us the area between two functions, \( f(x) \) and \( g(x) \), from x=a to x=b.
• Find bounds: In this example, we're looking between \( x = 0 \) and \( x = 2 \). These are our limits of integration.
• Integrate the difference: Substitute \( f(x) = -x^2 + 2x + 3 \) and \( g(x) = -x + 3 \) into the integral to get \[-x^2 + 3x\].
• Calculate the area: The definite integral \[ \left[-\frac{1}{3}x^3 + \frac{3}{2}x^2 \right]_0^2 \] is evaluated to find the solution.

Approaching problems through this structured method assures accuracy in solving for areas under, above, or in-between functions.

Area Between Curves

Finding the area between curves is an application of integration that allows us to calculate the space enclosed by two functions. To illustrate:

• Understand the setup: The problem asks for the area between \( f(x) = -x^2 + 2x + 3 \) and \( g(x) = -x+3 \), bound by vertical lines \( x = 0 \) and \( x = 2 \).
• Identify intersections: Already found as \( x = 0 \) (important for bounding the area).
• Apply calculations: By integrating the difference \( f(x) - g(x) \), you acquire the net area between two graphs.
• Consider limits: Only evaluate the integral within the prescribed limits \( a \) and \( b \), ensuring to consider only relevant segments.

The correct use of these steps results in finding an area of \( \frac{16}{3} \), a critical result achievable through a clear understanding of integration and graph behavior.