Problem 10 - Calculus: Early Transcendentals

Question

(a) Find the average value of \( f \) on the given interval. (b) Find \( c \) such that \( f_{ave} = f(c) \). (c) Sketch the graph of \( f \) and a rectangle whose area is the same as the area under the graph of \( f \). \( f(x) = \dfrac{1}{x} \) , \( [1, 3] \)

Step-by-Step Solution

Answer

(a) \( \frac{1}{2} \ln(3) \); (b) \( c = \frac{2}{\ln(3)} \).

Step 1: Calculate the Average Value of the Function

The average value of a function \( f \) on the interval \( [a, b] \) is given by the formula \[ f_{ave} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] Here, \( f(x) = \frac{1}{x} \), \( a = 1 \), and \( b = 3 \). Substitute these into the formula: \[ f_{ave} = \frac{1}{3-1} \int_{1}^{3} \frac{1}{x} \, dx = \frac{1}{2} \int_{1}^{3} \frac{1}{x} \, dx. \]

Step 2: Evaluate the Integral

Evaluate the integral \( \int_{1}^{3} \frac{1}{x} \, dx \). The integral of \( \frac{1}{x} \) is \( \ln|x| \): \[ \int_{1}^{3} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{3} = \ln(3) - \ln(1). \] Since \( \ln(1) = 0 \), we have \( \ln(3) \).

Step 3: Compute the Average Value

Substitute the result from Step 2 into the average value calculation: \[ f_{ave} = \frac{1}{2} \ln(3). \] This is the average value of \( f(x) \) on the interval \([1, 3]\).

Step 4: Find \( c \) such that \( f_{ave} = f(c) \)

We need \( f_{ave} = f(c) \), so \( \frac{1}{2} \ln(3) = \frac{1}{c} \). Solving for \( c \), we get: \[ c = \frac{2}{\ln(3)}. \]

Step 5: Sketch the Graph and the Rectangle

First, sketch the graph of \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 3 \). Next, draw a rectangle with height \( f_{ave} = \frac{1}{2} \ln(3) \) and base from \( x = 1 \) to \( x = 3 \). The area of the rectangle should equal the area under the graph, confirming \( 2 \times \frac{1}{2} \ln(3) = \ln(3) \).

Key Concepts

Integration

Integration is a fundamental concept in calculus, representing the process of finding the area under a curve. In the context of this exercise, it helps us determine the average value of a function. Think of integration as adding up infinitely many, infinitely small quantities. When you integrate a function over an interval, you are essentially summing up the values of the function over that range.

• The integral of a function gives the total accumulation of its values across the specified interval.
• It provides a way to find areas under curves, a crucial aspect in calculus.
• The notation for the integral of a function \( f(x) \) from \( a \) to \( b \) is \( \int_{a}^{b} f(x) \, dx \).

In our case, to find the average value of a function \( f(x) = \frac{1}{x} \) over the interval \([1, 3]\), we calculate:
\[ f_{ave} = \frac{1}{3-1} \int_{1}^{3} \frac{1}{x} \, dx = \frac{1}{2} \int_{1}^{3} \frac{1}{x} \, dx. \]
This tells us that the problem of finding the average value over \([1, 3]\) is simplified to evaluating an integral.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. This function is important in many areas of mathematics, including calculus and complex analysis.

• The natural logarithm is the inverse function of the exponential function.
• It helps in simplifying and solving logarithmic equations and integrating functions.
• In integration, the integral of \( \frac{1}{x} \) is \( \ln|x| \).

In step 2 of the solution, we evaluate the integral \( \int_{1}^{3} \frac{1}{x} \, dx \). Since the integral of \( \frac{1}{x} \) is \( \ln|x| \), it becomes:
\[ \left[ \ln|x| \right]_{1}^{3} = \ln(3) - \ln(1). \]
Knowing that \( \ln(1) = 0 \), we find the integration result to be \( \ln(3) \), essential for computing the average value.

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals provides a crucial link between the average value of a function and a particular value of the function within the given interval. It says that there exists at least one point \( c \) in the interval \([a, b]\) such that the integral of the function over \([a, b]\) is equal to \( f(c) \times (b-a) \).

• This theorem emphasizes that for continuous functions, the average value is not merely theoretical but actually occurs at some point within the interval.
• It helps in finding specific values that represent the behavior of a function averaged over its interval.
• The condition \( f_{ave} = f(c) \) helps identify where this average actually holds in the form of a rectangle's height on the graph.

In step 4, we use this theorem to find the value \( c \), where \( f(c) = \frac{1}{c} \) is equal to the average value \( f_{ave} = \frac{1}{2} \ln(3) \). Solving for \( c \), we determine:
\[ c = \frac{2}{\ln(3)}. \]
This shows that around \( x = c \), the function's value exactly equals the average value over the interval \([1, 3]\).