Problem 11 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=\ln (x) \quad I=[1, e] $$

Step-by-Step Solution

Answer

The average value of \( \ln(x) \) over \([1, e]\) is \( \frac{1}{e-1} \).

Step 1: Define the function and interval

We need to find the average value of the function \( f(x) = \ln(x) \) over the interval \( I = [1, e] \). The function is \( \ln(x) \) and the interval is from 1 to \( e \), where \( e \approx 2.718 \) is the base of the natural logarithm.

Step 2: Write the formula for the average value

The formula for the average value of a function \( f(x) \) over the interval \( [a, b] \) is given by: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. In this case, \( a = 1 \) and \( b = e \), so the formula becomes \[ \frac{1}{e-1} \int_{1}^{e} \ln(x) \, dx \].

Step 3: Set up the definite integral

We now need to evaluate the definite integral \( \int_{1}^{e} \ln(x) \, dx \). Use integration by parts, where you let \( u = \ln(x) \) and \( dv = dx \). Therefore, \( du = \frac{1}{x} \, dx \) and \( v = x \).

Step 4: Apply integration by parts

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), calculate \[ \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - \int 1 \, dx = x \ln(x) - x + C. \] Evaluate this from 1 to \( e \).

Step 5: Evaluate the integral from 1 to e

Substitute the limits of integration into the expression \( x \ln(x) - x \):\[ \left[ x \ln(x) - x \right]_{1}^{e} = \left( e \ln(e) - e \right) - \left( 1 \ln(1) - 1 \right). \] This simplifies to \( (e - e) - (0 - 1) = 0 + 1 = 1 \).

Step 6: Calculate the average value

Substitute the result of the integral and the limits into the average value formula: \[ \frac{1}{e-1} \cdot 1 = \frac{1}{e-1}. \] This is the average value of \( \ln(x) \) on the interval \([1, e]\).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals that are products of two functions. It is particularly useful when elementary methods such as substitution do not easily apply. The formula for integration by parts is derived from the product rule for differentiation and is given by:\[\int u \, dv = uv - \int v \, du\]Here’s how to apply it:

• Identify which part of your function will be \( u \) and which part will be \( dv \). A common choice is to let \( u \) be a logarithmic or polynomial function, because its derivative, \( du \), tends to simplify the integral.
• Let \( dv \) be the remaining part of the function. Find \( du \) by differentiating \( u \), and \( v \) by integrating \( dv \).
• Substitute these into the integration by parts formula and solve the resulting simpler integral, \( \int v \, du \).

In our exercise, \( u = \ln(x) \) and \( dv = dx \). Differentiating \( u \), we get \( du = \frac{1}{x} \, dx \), and integrating \( dv \), we get \( v = x \). This approach helps solve the integral \( \int \ln(x) \, dx \) by breaking it into manageable parts.

Definite Integral

The definite integral represents the area under a curve between two points on a graph. It is crucial in problems involving average values, such as the one given in the exercise. The definite integral is denoted by:\[\int_{a}^{b} f(x) \, dx\]

• The limits \( a \) and \( b \) are the lower and upper bounds of integration, respectively.
• Evaluating the definite integral involves finding an antiderivative of \( f(x) \), substituting the upper limit into it, then subtracting the result of substituting the lower limit.
• The result gives the net area between the function curve \( f(x) \) and the x-axis from \( a \) to \( b \).

In our case, the function \( \ln(x) \) was integrated from 1 to \( e \). After applying integration by parts, we evaluated it as \([ x \ln(x) - x ]\) from 1 to \( e \), finding the net area to be 1.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental function in calculus with many unique properties. It is the inverse of the exponential function \( e^x \).

• The natural logarithm is only defined for positive numbers. It answers the question, "To what power must \( e \) be raised to get the number \( x \)?"
• Its derivative is \( \frac{1}{x} \), which is often used in the integration by parts technique, as seen in our exercise.
• \( \ln(1) \) is 0 because \( e^0 = 1 \), and \( \ln(e) \) is 1 because \( e^1 = e \).

In the exercise, understanding the properties of \( \ln(x) \) helped evaluate the definite integral and determine how the function behaves over the specified interval \([1, e]\). This foundation is essential in both integration by parts and the calculation of definite integrals.