Problem 75 - Thomas Calculus

Question

a. Show that \(\int \ln x d x=x \ln x-x+C\) b. Find the average value of \(\ln x\) over \([1, e]\)

Step-by-Step Solution

Answer

a. Use integration by parts to show: \( \int \ln x \, dx = x \ln x - x + C \). b. Average value of \( \ln x \) over \([1, e]\) is \( \frac{1}{e-1} \).

Step 1: Use Integration by Parts Formula

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). For the integral \( \int \ln x \, dx \), let \( u = \ln x \) and \( dv = dx \). This choice means that \( du = \frac{1}{x} \, dx \) and \( v = x \), after integrating \( dv \).

Step 2: Apply Integration by Parts

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx. \] Simplify the integral to \( \int 1 \, dx \), which is equal to \( x \).

Step 3: Simplify the Expression

Substituting \( \int 1 \, dx = x \) into the equation, we have \[ \int \ln x \, dx = x \ln x - x. \] Thus, \( \int \ln x \, dx = x \ln x - x + C \), which demonstrates the initial given equation.

Step 4: Set Up the Average Value Formula

The formula for the average value of a function \( f(x) \) over an interval \([a, b]\) is \( \frac{1}{b-a} \int_a^b f(x) \, dx \). For \( f(x) = \ln x \) on \([1, e]\), this becomes \( \frac{1}{e-1} \int_1^e \ln x \, dx \).

Step 5: Compute the Definite Integral

Using the result from part (a), find \( \int_1^e \ln x \, dx = \left[ x \ln x - x \right]_1^e \). Calculate this as \( \left[e \ln e - e\right] - \left[1 \ln 1 - 1\right] = \left[e - e\right] - \left[0 - 1\right] = 1 \).

Step 6: Calculate the Average Value

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

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that calculates the net 'area' under a curve, between two boundaries on the x-axis. It can be thought of as adding up infinitely many, infinitely small products of height (the function value) and width (a tiny piece of the x-axis).
To evaluate a definite integral, you find the antiderivative of the function, then calculate the difference of its values at the upper and lower limits.

• For example, given the function \( f(x) \, = \, \ln x \, \,\) on the interval \([1, e]\), the definite integral is calculated by finding \( \left[ x \ln x - x \right]_1^e\).
• This step involves evaluating the expression \( x \ln x - x \) at both 1 and \( e \), and then subtracting the results.
• The definite integral \( \int_1^e \ln x \, dx \) results in the value 1, after applying this process.

Logarithmic Functions

Logarithmic functions are inverses of exponential functions and are represented as \( y = \ln x \), where \( \ln \) denotes the natural logarithm with base \( e \) (Euler's number, approximately 2.718).
These functions have unique characteristics that make them useful in various applications.

• The derivative of \( \ln x \) is \[ \frac{d}{dx} \ln x = \frac{1}{x}, \] which plays a key role in calculus, especially in integration techniques like integration by parts.
• Logarithmic functions grow slowly compared to polynomial and exponential functions, making them ideal for modeling situations where growth decreases over time, such as in certain economic models.
• In integration by parts, choosing \( u = \ln x \) simplifies the derivation as its derivative is straightforward, aiding in finding its integral.

Average Value of a Function

The average value of a function over a certain interval provides a single number that represents the 'average' output over that interval. This is particularly useful when you want to simplify the analysis of the function's behavior over time or space.
The formula for the average value of a function \( f(x) \) over the interval \( [a, b] \) is \[ \frac{1}{b-a} \int_a^b f(x) \, dx. \]

• This formula divides the integral of the function over the interval by the length of the interval, thus averaging the total "accumulated" value.
• For example, for the function \( \ln x \) over the interval \( [1, e] \), the calculation resulted in \( \frac{1}{e-1} \) after evaluating the integral previously found to be 1.
• This shows that despite the complexity of the function, the average value over this interval is surprisingly simple and can be easily used for further analysis.