Problem 24 - Applied Calculus

Question

Find the average value of each function over the given interval. \(f(x)=\frac{1}{x}\) on \([1, c],\) where \(c\) is a constant \((c>1)\)

Step-by-Step Solution

Answer

The average value is \( \frac{\ln(c)}{c-1} \).

Step 1: Write the Formula for Average Value of a Function

The average value of a continuous function \( f(x) \) over an interval \([a, b]\) can be calculated using the formula: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \,dx \] Here, we need to find the average value of \( f(x) = \frac{1}{x} \) over the interval \([1, c]\). In this case, \(a = 1\) and \(b = c\).

Step 2: Set Up the Integral for the Function

For \( f(x) = \frac{1}{x} \) over the interval \([1, c]\), set up the integral:\[ \int_{1}^{c} \frac{1}{x} \, dx \]This integral will be used to calculate the average value.

Step 3: Evaluate the Integral

The integral of \( \frac{1}{x} \) is \( \ln|x| \). Evaluate the definite integral:\[ \int_{1}^{c} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{c} = \ln|c| - \ln|1| \]Since \( \ln|1| = 0 \), it simplifies to:\[ \ln(c) \]

Step 4: Calculate the Average Value

Substitute the result from the integral back into the average value formula:\[ f_{avg} = \frac{1}{c-1} (\ln(c) - \ln(1)) = \frac{1}{c-1} \ln(c) \] This is the average value of the function \( f(x) = \frac{1}{x} \) over the interval \([1, c]\).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus. It represents the accumulation of quantities, which could be areas under curves or other physical quantities. The process of finding definite integrals is often called integration.
In mathematics, a definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is a function defined over the interval \([a, b]\). This notation means we're calculating the total area between the curve \( f(x) \), the x-axis, and between the limits \( x = a \) and \( x = b \).
Some key points about definite integrals are:

• They can yield positive or negative values, depending on whether the function lies above or below the x-axis.
• The definite integral between limits \( a \) and \( b \) gives a specific number, unlike indefinite integrals which include a constant of integration.
• The result can represent various real-world quantities like total distance traveled, volume, and more.

In the context of calculating the average value of a function, the definite integral helps determine the total value of the function across an interval. This total is then divided by the length of the interval to find the average.

Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm has many unique and useful properties in calculus and analysis.
Some insight on natural logarithms include:

• They are used to solve exponential equations where the base is \( e \).
• The natural logarithm of 1 is always zero: \( \ln(1)=0 \).
• For any positive number \( x \), \( \ln(x) \) is the exponent to which \( e \) must be raised to produce \( x \).
• Natural logarithms have important applications in many areas such as compound interest, growth processes, and more.

In the given exercise, the integral of \( \frac{1}{x} \) results in a natural logarithm, \( \ln|x| \). This transformation is crucial towards obtaining the average value of the function over the specified interval.

Continuous Function

A continuous function is a pivotal concept in calculus that ensures the function does not have any abrupt changes or discontinuities over its domain. In simple terms, a continuous function is one you can draw without lifting your pen.
Key characteristics of continuous functions include:

• No breaks, jumps, or holes: The function is unbroken across its domain.
• If a function is continuous over an interval \([a, b]\), then the definite integral of the function can be successfully evaluated over that interval.
• Limits exist at every point, and the limit is equal to the function's value at each of those points.

In our exercise, the function \( f(x) = \frac{1}{x} \) is continuous on \((0, \infty)\), which includes the interval \([1, c]\) that we are concerned with. This property is crucial because the formula for average value relies on the function being continuous over the interval of integration.