Problem 56 - Calculus and its applications

Question

Evaluate. Verify that for any positive integer \(n\) $$ \int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x $$

Step-by-Step Solution

Answer

The formula is verified by using integration by parts and simplifying the result.

Step 1: Understand the Problem

We need to verify that the integral of a natural logarithm raised to a power satisfies a certain formula. Specifically, for a positive integer \( n \), we are given:\[ \int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x \]

Step 2: Use Integration by Parts

The integral \( \int(\ln x)^n \, dx \) is best approached using integration by parts. Let \( u = (\ln x)^n \) and \( dv = dx \). Then, \( du = n (\ln x)^{n-1} \cdot \frac{1}{x} \cdot dx \) and \( v = x \).

Step 3: Apply the Integration by Parts Formula

Integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). Substituting the parts:\[ \int(\ln x)^n \, dx = x (\ln x)^n - \int x \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]Simplify the integral:\[ = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]

Step 4: Verify by Simplification

After applying integration by parts and simplifying, the result we have matches the formula given in the problem. Thus, the verification step shows that:\[ \int (\ln x)^n \, dx = x(\ln x)^n - n \int (\ln x)^{n-1} \, dx \] is indeed correct.

Key Concepts

Integral of Natural Logarithm

When dealing with integrals, understanding the integral of the natural logarithm is crucial. The exercise involves the integral \( \int (\ln x)^n \, dx \). This can be a bit challenging at first because natural logarithms are not polynomial or simple power functions. However, using integration by parts simplifies the process. Integration by parts is a technique derived from the product rule for differentiation, allowing us to tackle integrals of products of functions.

Using integration by parts, we let \( u = (\ln x)^n \) and \( dv = dx \). Through differentiation, \( du = n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \), and because \( dv = dx \), then \( v = x \). When we apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \), it transforms our problem into a simpler form. The process results in breaking down the original integral into a manageable expression that adheres to the given formula in the problem statement.

Understanding this aspect makes the seemingly complex integral of a natural logarithm more approachable, especially when verifying formulas involving these kinds of functions.

Logarithmic Functions

Logarithms are essential in various branches of mathematics, particularly calculus. The natural logarithm, expressed as \( \ln x \), has unique properties.

Some key properties include:

• The natural logarithm is the inverse of the exponential function \( e^x \).
• It is defined as the area under the curve \( y = \frac{1}{x} \) from 1 to \( x \).
• The derivative of \( \ln x \) is \( \frac{1}{x} \) and its integral \( \int \ln x \, dx \) requires integration by parts due to its non-trivial form.

Understanding these properties is vital for any calculus student because they provide a foundation for solving more complex calculus problems involving logarithms, such as integration by parts used in this problem.

Logarithms, particularly natural logarithms, appear across various applications in science and engineering due to their relationship with exponential growth and decay. Recognizing and using these properties effectively will ease the process of calculating integrals that include logarithmic functions.

Calculus Proof Techniques

Proof techniques are the backbone of calculus, ensuring that derived results are valid and reliable. In the exercise, we verified an integral involving powers of the natural logarithm by applying a systematic method.

Here are some useful proof techniques often applied in calculus:

• Integration by Parts: A sophisticated method, as demonstrated in the exercise to simplify complex integrals.
• Substitution: Changing variables to simplify the integration process.
• Comparison: Used for definite integrals, comparing a function to a simple bounding function.
• Induction: Especially useful when dealing with properties involving integers, such as powers of a logarithm.

The focus in this exercise was to use integration by parts, simplifying a more involved logarithm-based integral into a form that makes it clear and manageable.

Mastering these techniques not only aids verification of results but also builds confidence in solving various calculus problems. They are tools in a mathematician's toolbox, each helpful in different scenarios, ensuring the integration and other processes are done accurately.