Problem 16 - Calculus Single Variable

Question

In each of Exercises \(13-16,\) a function \(f\) and an interval \(I=[a, b]\) are given. Find a number \(c\) in \((a, b)\) for which \(f(c)\) is the average value of \(f\) on \(I\). $$ f(x)=\ln (x) \quad I=[1, e] $$

Step-by-Step Solution

Answer

The number \(c\) is \(e^{\frac{1}{e-1}}\).

Step 1: Understand the Problem

We need to find a number \(c\) within the interval \((1, e)\) such that the function value \(f(c)\) equals the average value of \(f(x) = \ln(x)\) over the interval \([1, e]\).

Step 2: Find the Average Value of the Function

The average value \(f_{avg}\) of a continuous function \(f(x)\) over the interval \([a, b]\) is given by \(f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx\). For our function, it simplifies to \(f_{avg} = \frac{1}{e-1} \int_1^e \ln(x) \, dx\).

Step 3: Compute the Integral

First, we solve the integral: \(\int_1^e \ln(x) \, dx\). To do this, we use integration by parts: let \(u = \ln(x)\) and \(dv = dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = x\). The integration by parts formula \(\int u \, dv = uv - \int v \, du\) gives:\[ \int \ln(x) \, dx = x \ln(x) - \int x \frac{1}{x} \, dx = x \ln(x) - x. \]Evaluating from 1 to \(e\), we get:\[ \left[ x \ln(x) - x \right]_1^e = \left[e \ln(e) - e \right] - \left[1 \ln(1) - 1 \right]= (e - e) - (0 - 1) = 1.\]

Step 4: Calculate the Average Value

Now substitute back to find \(f_{avg}\) using the integral result:\[ f_{avg} = \frac{1}{e-1}(1) = \frac{1}{e-1}. \]

Step 5: Solve for \(c\) such that \(f(c) = f_{avg}\)

We need \(f(c) = \ln(c) = \frac{1}{e-1}\). Solve for \(c\) by exponentiating both sides:\[ \ln(c) = \frac{1}{e-1} \Rightarrow c = e^{\frac{1}{e-1}}. \]

Step 6: Verify the Solution

Verify that \(c = e^{\frac{1}{e-1}}\) lies within the interval \((1, e)\). Since \(\frac{1}{e-1} > 0\), \(c = e^{\frac{1}{e-1}} > 1\), and because \(e^{\frac{1}{e-1}} < e\), \(c\) indeed is in the interval. Thus, \(c = e^{\frac{1}{e-1}}\) is correct.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique often used to solve integrals that involve the product of two functions. It's especially handy when direct integration is complicated or impossible.
This method stems from the product rule of differentiation and can be formulated as:

• Choose a function to differentiate (usually the polynomial or easily differentiable function).
• Pick a function to integrate (often an exponential, logarithmic, or trigonometric function).

Given two functions, say \(u(x)\) and \(v(x)\), the integration by parts formula is:\[\int u \, dv = uv - \int v \, du.\]In the solution, we applied integration by parts to integrate \(\ln(x)\). We chose \(u = \ln(x)\) and \(dv = dx\), resulting in \(du = \frac{1}{x} \, dx\) and \(v = x\). After integration, we evaluated at the limits to find the definite integral result.

Average Value of a Function

The average value of a function over a specific interval provides a single number that represents the overall "height" of the function on that interval.
This concept is similar to finding the average of a set of numbers. In calculus, it's given by:

• Identify the interval \[ [a, b]. \]
• Use the formula: \[f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx.\]

In the example, the interval \[ [1, e] \] was used. We calculated the average value of the function \(f(x) = \ln(x)\) across this interval. By computing the definite integral \(\int_1^e \ln(x) \, dx\) and dividing by the length of the interval \( e-1 \), we obtained the average value \(f_{\text{avg}} = \frac{1}{e-1}\).

Definite Integral

A definite integral is a fundamental concept in calculus that allows us to calculate the accumulation of quantities, such as areas under curves.
It is expressed as:

• The symbol \(\int\) denotes integration.
• The limits \([a, b]\) are the interval's endpoints.

When calculating a definite integral, we are finding the net area under the curve from point \(a\) to point \(b\) on a graph of the function. The process involves:

• Calculating the antiderivative (or integral) of the function.
• Evaluating this antiderivative at the upper bound \((b)\) and lower bound \((a)\).
• Subtracting these values to find the net area.

In the given solution, we computed \(\int_1^e \ln(x) \, dx\) by first determining the antiderivative and then evaluating it from \(1\) to \(e\) to find the result of \(1\).

Continuous Function

A continuous function is a crucial concept in calculus, representing functions without breaks, jumps, or holes in their graphs.
Continuity ensures that the function behaves predictably and has no abrupt changes. For a function \(f(x)\) to be continuous over an interval \[ [a, b], \]

• \(f(x)\) must be defined for every point between \(a\) and \(b\).
• As \(x\) approaches any point \(c\) in \([a, b],\)\(f(x)\) should smoothly approach \(f(c)\).

Continuous functions are essential when finding average values or evaluating integrals, as they guarantee that all necessary computations can take place without issue. In this example, \(f(x) = \ln(x)\) is continuous on \([1, e]\), ensuring that calculating its average value and definite integral over this interval are valid and straightforward.