Problem 32 - Calculus Early Transcendentals

Question

The elevation of a path is given by \(f(x)=x^{3}-5 x^{2}+30,\) where \(x\) measures horizontal distances. Draw a graph of the elevation function and find its average value, for \(0 \leq x \leq 4\).

Step-by-Step Solution

Answer

Answer: The average value of the elevation function over the interval is \(\frac{8}{3}\).

Step 1: Identify the function

The function representing the elevation of the path is given by \(f(x) = x^3 - 5x^2 + 30\), for horizontal distances \(x\).

Step 2: Plot the points for the graph

Calculate the elevation at representative x-values and plot the points on graph paper: $$ f(0) = (0)^3 - 5(0)^2 + 30 = 30 \\ f(1) = (1)^3 - 5(1)^2 + 30 = 26 \\ f(2) = (2)^3 - 5(2)^2 + 30 = 14 \\ f(3) = (3)^3 - 5(3)^2 + 30 = 0 \\ f(4) = (4)^3 - 5(4)^2 + 30 = -14 $$ Plot the points \((0, 30), (1, 26), (2, 14), (3, 0), (4, -14)\), and connect them to create the graph of the elevation function.

Step 3: Find the average value of the function over the interval

The average value of the function over the interval \(0 \leq x \leq 4\) is given by: $$ \frac{1}{4 - 0} \int_{0}^{4} (x^3 - 5x^2 + 30) dx $$

Step 4: Integrate the function

Integrate the function with respect to \(x\): $$ \int_{0}^{4} (x^3 - 5x^2 + 30) dx = \left[ \frac{1}{4} x^4 - \frac{5}{3} x^3 + 30x \right]_0^4 $$

Step 5: Evaluate and simplify

Evaluate the integral and simplify: $$ \frac{1}{4} \left[ \left( \frac{1}{4} (4)^4 - \frac{5}{3} (4)^3 + 30(4) \right) - \left( \frac{1}{4} (0)^4 - \frac{5}{3} (0)^3 + 30(0) \right) \right] = \frac{1}{4} \left[ \left( 64 - \frac{320}{3} + 120 \right) - 0 \right] = \frac{8}{3} $$ The average value of the elevation function \(f(x) = x^3 - 5x^2 + 30\) over the interval \(0 \leq x \leq 4\) is \(\frac{8}{3}\).

Key Concepts

Integration

Integration is a key concept in calculus that allows us to find areas under curves and the accumulation of quantities. In this exercise, integration is used to determine the average value of the function over a specific interval. The function, \( f(x) = x^3 - 5x^2 + 30 \), is integrated over the interval from \(0\) to \(4\).
To find the average value, we need to integrate the function:

• First, set up the integral for \(\int_{0}^{4}(x^3 - 5x^2 + 30)\, dx\).
• Apply the power rule for integration, resulting in the antiderivative \[\frac{1}{4}x^4 - \frac{5}{3}x^3 + 30x.\]
• Evaluate this antiderivative at the boundaries of the interval \(0\) and \(4\).

After calculation, the definite integral evaluates to the quantity
\[\left(64 - \frac{320}{3} + 120\right)\].
Finally, to find the average, divide this result by the length of the interval (\(4 - 0\)), resulting in \(\frac{8}{3}\). The average represents the mean elevation over the distance.

Plotting Graphs

Plotting graphs is a fundamental skill in understanding functions visually. It provides a clear view of the behavior of the function at different points. In this case, the function \( f(x) = x^3 - 5x^2 + 30 \) was plotted from \( x = 0 \) to \( x = 4 \).
Here's how you can visualize the function effectively:

• Calculate the function's value at strategic points: \(f(0) = 30\), \(f(1) = 26\), \(f(2) = 14\), \(f(3) = 0\), and \(f(4) = -14\).
• Plot these points on a graph on the x-y plane with the x-values on the horizontal axis.
• Join the points with a smooth curve to form the graph of the function.

Seeing these points and the curve connecting them enables you to discern the function's general shape and changes. Particularly, it reveals how the elevation decreases from \(f(0) = 30\) to \(f(4) = -14\).
This visual representation makes it easier to grasp function behavior and can aid in predicting and understanding its average value.

Polynomial Functions

Polynomial functions are mathematical expressions involving variables raised to positive integer powers. The function \( f(x) = x^3 - 5x^2 + 30 \) is a polynomial of degree 3, also known as a cubic polynomial.
Key features of polynomial functions include:

• They have a smooth and continuous graph, with no breaks or gaps.
• The degree of the polynomial (the highest power of the variable) largely determines the graph's shape.
• The leading term (\(x^3\) in this case) determines the end behavior of the graph.

In this exercise, the variable \(x\) represents horizontal distance, and the function describes the path's elevation. The cubic nature implies the graph will have at most two turning points – locations where the graph changes direction. Understanding these characteristics helps predict how the function behaves beyond just given values.
Polynomials like this one are common in real-world applications, such as modeling physical paths, economic trends, and natural phenomena.