Problem 71 - 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. $$ f(x)=x^{2}+4 / x a=2 \quad b=6 $$

Step-by-Step Solution

Answer

\(f_{\text{avg}} = \frac{52}{3} + \ln 3\); find \(c\) using numerical methods in \((2,6)\).

Step 1: Set Up the Integral for Average Value

To find the average value of the function over the interval \([a, b]\), we use the formula for the average value of a function:\[ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \]In this case, the function is \(f(x) = x^2 + \frac{4}{x}\), and the interval is \([2, 6]\). Therefore, we plug these into the formula:\[ f_{\text{avg}} = \frac{1}{6-2} \int_2^6 \left(x^2 + \frac{4}{x}\right) dx \]

Step 2: Compute the Integral

Evaluate the integral \( \int_2^6 \left(x^2 + \frac{4}{x}\right) dx \):Compute separately:\[ \int x^2 \, dx = \frac{x^3}{3} \]\[ \int \frac{4}{x} \, dx = 4 \ln|x| \]At \([2, 6]\), this integral becomes:\[ \left[ \frac{x^3}{3} + 4 \ln|x| \right]_2^6 = \left(\frac{6^3}{3} + 4 \ln|6|\right) - \left(\frac{2^3}{3} + 4 \ln|2|\right) \]

Step 3: Evaluate the Integral Expression

Calculate the values:\[ \frac{6^3}{3} + 4 \ln|6| = 72 + 4 \ln 6 \]\[ \frac{2^3}{3} + 4 \ln|2| = \frac{8}{3} + 4 \ln 2 \]Subtract these results:\[ \left(72 + 4 \ln 6\right) - \left(\frac{8}{3} + 4 \ln 2\right) = \frac{208}{3} + 4 (\ln 3) \]

Step 4: Compute the Average Value

Plug the result from the integral into the average value formula:\[ f_{\text{avg}} = \frac{1}{4} \left(\frac{208}{3} + 4 \ln 3\right) = \frac{52}{3} + \ln 3 \]Thus, the average value \( f_{\text{avg}} \) is \( \frac{52}{3} + \ln 3 \).

Step 5: Find the Value of c

We need \(c\) such that \(f(c) = f_{\text{avg}}\). Solve \(f(c) = c^2 + \frac{4}{c} = \frac{52}{3} + \ln 3\).This equation requires numerical or graphical methods to find \(c\) since it's not easily solvable algebraically for exact values. Approximate \(c\) by testing values in \((2, 6)\) to find where \(f(c)\) is close to this value. This can be done using graphing technology.

Step 6: Explain the Geometric Meaning

According to the Mean Value Theorem for Integrals, there exists a \(c\) in \((2, 6)\) such that the area under \(f(x)\) from \(2\) to \(6\) is equal to the area of the rectangle with height \(f_{\text{avg}}\) and base \(6-2=4\).Graph \(f(x)\) and a horizontal line at \(f_{\text{avg}}\). The point where this line intersects \(f(x)\) within the interval \([2, 6]\) gives \(c\).

Key Concepts

Average Value of a Function

When learning about the average value of a function on an interval, one is essentially discovering the typical or mean value that the function achieves over that interval. The concept is much like finding the average of a set of numbers, but instead, it's applied over a continuous range. The mathematical expression for the average value of a function \( f(x) \) on an interval \( [a, b] \) is given by: \[ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \] This involves computing the definite integral over the interval \([a, b]\), which calculates the area under the curve of \(f(x)\).

• The formula helps us determine a single value that represents the function's average behavior over that interval.
• This average value provides insight into the function's behavior without needing to study every value it takes individually over \([a, b]\).

In practice, this average value can be used to predict outcomes or understand systems modeled by functions, and is computed with relative ease using the integral formula. This understanding of averages encourages the appreciation of the utility of calculus in summarizing complex sets of information.

Definite Integrals

Definite integrals are a fundamental concept in calculus with a variety of practical applications, crucial for calculating areas under curves. Computing a definite integral of a function \( f(x) \) over an interval \( [a, b] \) is expressed as: \[ \int_a^b f(x) \, dx \].

• The result of a definite integral represents the net area between the function and the x-axis across the given interval.
• This net area accounts for both, regions where the function is above the axis (positive area) and below the axis (negative area).

In the context of the exercise where \( f(x) = x^2 + \frac{4}{x} \), evaluating the definite integral from 2 to 6 requires finding the antiderivatives of each term, plugging in the limits, and calculating the difference. These values provide insights into the function's overall behavior and facilitate calculations like average values. Understanding how definite integrals uncover the area under curves helps ground abstract mathematical concepts in tangible geometric interpretations.

Geometric Interpretation of Calculus

The geometric interpretation of calculus focuses on understanding shapes and areas using calculus tools like integration. Calculus aids in visualizing problems, and the Mean Value Theorem for Integrals particularly underscores this connection. The theorem states that, given a continuous function on a closed interval \([a, b]\), there exists at least one point \(c\) within that interval where the function equals its average value: \( f(c) = f_{\text{avg}} \).

• This highlights that the area under the curve can be equivalent to the area of a rectangle with width equal to the interval and height equal to the average value.
• Graphically, this emerges as a horizontal line intersecting the curve at least once within \([a, b]\).

Visualizing the situation using a graph helps concretize abstract concepts, showing precisely where and how average value coincides in the function's range. By graphing \(f(x)\) and the line \(f_{\text{avg}}\), students can easily see the intersection, which not only proves mathematical theorems but also enhances their intuitive understanding of calculus and its applications in determining shapes and sizes geometrically.