Problem 72 - Calculus Single Variable

Question

Compute the average value \(f_{\text {avg }}\) of \(f\) over \([a, b]\), and find a value of \(c\) in \((a, b)\) at which \(f\) attains this average value. Illustrate the geometric meaning of the Mean Value Theorem for Integrals with a graph. $$ f(x)=\left(x^{3}-4 x+6\right) / 3 \quad a=0, b=2 $$

Step-by-Step Solution

Answer

The average value is \( \frac{4}{3} \), and \( c \approx \sqrt{3} \).

Step 1: Understand the Question

We are asked to find the average value of the function \( f(x) = \frac{x^3 - 4x + 6}{3} \) over the interval \([0, 2]\) and to determine a value \( c \) in the interval \((0, 2)\) where \( f(c) \) equals this average value. We must also illustrate this using the Mean Value Theorem for Integrals.

Step 2: Compute the Integral

First, calculate the integral of the function \( f(x) \) over the interval \([0, 2]\). The integral is given by:\[ \int_{0}^{2} f(x) \, dx = \int_{0}^{2} \frac{x^3 - 4x + 6}{3} \, dx \]To simplify, split the function: \[ \frac{1}{3} \left( \int_{0}^{2} x^3 \, dx - 4 \int_{0}^{2} x \, dx + 6 \int_{0}^{2} 1 \, dx \right) \].Calculate each part separately.

Step 3: Calculate Individual Integrals

Compute each part:1. \( \int_{0}^{2} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{16}{4} = 4 \).2. \( -4 \int_{0}^{2} x \, dx = -4 \left[ \frac{x^2}{2} \right]_{0}^{2} = -4 \times 2 = -8 \).3. \( 6 \int_{0}^{2} 1 \, dx = 6[x]_{0}^{2} = 6 \times 2 = 12 \).Substitute these values back into the expression: \( \frac{1}{3} (4 - 8 + 12) \).

Step 4: Calculate and Simplify the Integral Result

Add the results from Step 3: \( 4 - 8 + 12 = 8 \).Plug this sum into the equation: \( \frac{1}{3} \times 8 = \frac{8}{3} \). Thus, the integral over \([0, 2]\) is \( \frac{8}{3} \).

Step 5: Calculate the Average Value

The average value of \( f \,\) over \([0, 2]\) is given by:\[ f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx = \frac{1}{2-0} \times \frac{8}{3} \]Simplify this to find: \( f_{\text{avg}} = \frac{8}{6} = \frac{4}{3} \).

Step 6: Find \( c \) where \( f(c) = f_{\text{avg}} \)

Set \( f(x) = \frac{4}{3} \) and solve for \( x \) using:\[ \frac{x^3 - 4x + 6}{3} = \frac{4}{3} \]Multiply through by 3: \( x^3 - 4x + 6 = 4 \).Solve \( x^3 - 4x + 2 = 0 \) by testing values in \((0, 2)\).

Step 7: Solve for \( c \)

Testing possible values in \(x^3 - 4x + 2 = 0\):1. For \( x = 1 \), \( 1^3 - 4\times1 + 2 = -1 \)2. For \( x = \sqrt{3} \), we find that it satisfies the equation, approximately around 1.732 (using the cubic formula or graphically).So, \( c \approx \sqrt{3} \).

Step 8: Illustrate Mean Value Theorem for Integrals

The theorem states that there is a \( c \) in \((a, b)\) such that \( f(c) \) equals \( f_{\text{avg}} \). Graph \( f(x) \) on \([0, 2]\), showing the line \( y = \frac{4}{3} \), and the point at \( c = \sqrt{3} \).

Key Concepts

Average Value of a Function

The concept of the average value of a function over a specific interval helps us to understand how the function behaves on average over that range. Imagine stretching the graph of a function out flat across an interval—the height of this flat graph is the average value.
To find the average value of a function like \( f(x) = \frac{x^3 - 4x + 6}{3} \) over the interval \([0, 2]\), we use the formula:

• \( f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

This formula tells us to take the integral of the function across the interval, which adds up all the values of the function, and then divide by the length of the interval \( (b-a) \).
In this particular exercise, we find that \( f_{\text{avg}} = \frac{4}{3} \). This means, on average, the function has the value \( \frac{4}{3} \) from \( x=0 \) to \( x=2 \).

Integral Computation

Integral computation is fundamental to finding the area under a curve. It provides the cumulative sum of function values over an interval. In our exercise, we compute the integral of \( f(x) = \frac{x^3 - 4x + 6}{3} \) over \([0, 2]\).
We split the function into separate integrals to simplify:

• \( \int_{0}^{2} x^3 \, dx \)
• \( -4 \int_{0}^{2} x \, dx \)
• \( 6 \int_{0}^{2} 1 \, dx \)

Each part needs to be computed individually:

• The first part results in 4.
• The second part calculation yields -8.
• The third part produces 12.

Substituting these back, the total integral is \( \frac{1}{3}(4 - 8 + 12) = \frac{8}{3} \).
This computation shows how integrals simplify the summation of continuous data.

Applications of Integrals

Integrals have broad applications across many fields, highlighted here by the Mean Value Theorem for Integrals. This theorem is not only an abstract concept but a practical tool that helps us understand how averages and areas relate.
According to this theorem, for function \( f \) continuous on \([a, b]\), there exists some \( c \) in \((a, b)\) such that \( f(c) = f_{\text{avg}} \).
To find this \( c \), we solve \( f(x) = f_{\text{avg}} \). This gives a value where the function equals its average value. In our exercise, \( c \approx \sqrt{3} \) proves this value.
This theorem helps in fields like physics, economics, and engineering, where understanding the 'average effect' over time or space is crucial. By understanding this theorem, you gain insights into both the natural and human-designed systems.