Problem 143 - Thomas Calculus

Question

\begin{equation}\begin{array}{l}{\text { a. Show that } \int \ln x d x=x \ln x-x+C.} \\ {\text { b. Find the average value of } \ln x \text { over }[1, e].}\end{array}\end{equation}

Step-by-Step Solution

Answer

a. \(x \ln x - x + C \); b. \(\frac{1}{e-1}\).

Step 1: Integration by Parts Formula

To solve \( \int \ln x \, dx \), we use the integration by parts formula, which is \( \int u \, dv = uv - \int v \, du \). Here, if we set \( u = \ln x \), then \( du = \frac{1}{x} \, dx \). We choose \( dv = dx \), which makes \( v = x \).

Step 2: Apply Integration by Parts

Substitute into the integration by parts formula: \[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx. \] Simplifying, we get \[ \int \ln x \, dx = x \ln x - \int 1 \, dx. \]

Step 3: Simplify the Remaining Integral

The integral \( \int 1 \, dx \) simply evaluates to \( x \). Substituting back, we have: \[ \int \ln x \, dx = x \ln x - x + C. \] This confirms part (a) of the problem.

Step 4: Set Up Average Value Formula

For part (b), the average value \( f_{avg} \) of a function \( f(x) \) over an interval \([a, b]\) is given by \[ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx. \] In this case, \( f(x) = \ln x \), \( a = 1 \), and \( b = e \).

Step 5: Calculate the Average Value

Plugging into the formula gives \[ f_{avg} = \frac{1}{e-1} \int_1^e \ln x \, dx. \] We already know \( \int \ln x \, dx = x \ln x - x + C \). Thus, \[ \int_1^e \ln x \, dx = \left[ x \ln x - x \right]_1^e. \]

Step 6: Evaluating the Definite Integral

Evaluate \( \left[ x \ln x - x \right]_1^e \): - At \( x = e \), \( e \ln e - e = e \cdot 1 - e = 0 \).- At \( x = 1 \), \( 1 \ln 1 - 1 = 0 - 1 = -1 \).Therefore, the integral from \( 1 \) to \( e \) equals \( 0 - (-1) = 1 \).

Step 7: Calculate the Final Average Value

Substitute the result from Step 6 into the average value formula to find \( f_{avg} \): \[ f_{avg} = \frac{1}{e-1} \times 1 = \frac{1}{e-1}. \] This gives us the average value of \( \ln x \) over the interval \([1, e]\).

Key Concepts

Definite Integral

A definite integral is a key concept in calculus that helps us find the total amount of change over an interval. It is represented as \( \int_a^b f(x) \, dx \), where \(f(x)\) is the function we want to integrate over the interval \([a, b]\). Unlike indefinite integrals, definite integrals result in a specific number instead of a function plus a constant of integration \(C\). Definite integrals are used in a variety of applications, such as finding areas under curves, volumes of solids, and averages of functions on a closed interval.

When solving a definite integral, we often use the Fundamental Theorem of Calculus, which connects differentiation and integration. By evaluating the antiderivative or indefinite integral of a function at the boundaries \(a\) and \(b\), we subtract the values to obtain the definite integral:

• Find the antiderivative \(F(x)\) of \(f(x)\).
• Calculate \(F(b) - F(a)\).

For instance, when evaluating \( \int_1^e \ln x \, dx \), we substitute into the antiderivative \( x \ln x - x \) at \(e\) and \(1\) to find the net accumulated change over that interval.

Average Value of a Function

The average value of a function over a specific interval gives us a sense of the function's typical behavior within that range. It's like finding the average height in a group of people: it represents a middle ground. Mathematically, if you want to find the average value of a function \(f(x)\) from \(x = a\) to \(x = b\), you use the formula: \[f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx\]This formula signifies dividing the total accumulation of \(f(x)\) over the interval by the length of the interval \([a, b]\).

In our exercise, we calculated the average value of \( \ln x \) over the interval \([1, e]\) using this exact approach. By computing the definite integral \( \int_1^e \ln x \, dx = 1\), we then divided by \(e-1\) to normalize it over the length of the interval, resulting in an average value of \( \frac{1}{e-1} \). This indicates how the log function behaves on average from 1 to \(e\).

Natural Logarithm

The natural logarithm, denoted \( \ln x \), is a special logarithm with the base \(e\), where \(e\) is a mathematical constant approximately equal to 2.71828. It's a crucial function in many areas of mathematics due to its natural properties arising from calculus and exponential growth or decay.

The natural logarithm function has the fundamental property that \( \ln e = 1 \) because \(e^1 = e\). Additionally, logarithms turn multiplications into additions: \( \ln(xy) = \ln x + \ln y \). This can simplify complex calculations significantly.

In calculus, \( \ln x \) has the derivative \( \frac{d}{dx} \ln x = \frac{1}{x} \), making it widely applicable in solving integration problems and differential equations. Its integral, as derived in the exercise through integration by parts, is \( \int \ln x \, dx = x \ln x - x + C \). This result is important for understanding how a natural logarithm integrates over a continuous interval, helping resolve functions involving \( \ln x \) in practical applications.