Problem 68 - Applied Calculus

Question

Each equation follows from the integration by parts formula by replacing \(u\) by \(f(x)\) and \(v\) by a particular function. What is the function \(v ?\) \(\int f(x) \frac{1}{x} d x=f(x) \ln x-\int \ln x f^{\prime}(x) d x\)

Step-by-Step Solution

Answer

The function \( v \) is \( \ln x \).

Step 1: Identify Integration by Parts Formula

The integration by parts formula states that for functions \( u \) and \( dv \), \( \int u \, dv = uv - \int v \, du \). In this problem, \( u = f(x) \) and \( dv = \frac{1}{x} \, dx \).

Step 2: Determine \( v \) from \( dv \)

Given \( dv = \frac{1}{x} \, dx \), we find \( v \) by integrating \( dv \). Thus, \( v = \int \frac{1}{x} \, dx = \ln|x| \).

Step 3: Plug Values into Integration by Parts Formula

Insert the identified values into the integration by parts formula: \( \int f(x) \frac{1}{x} \, dx = f(x) \ln|x| - \int \ln|x| f'(x) \, dx \).

Step 4: Identify Function \( v \) in Given Equation

In the given equation, it is stated that \( \int f(x) \frac{1}{x} \, dx = f(x) \ln x - \int \ln x f'(x) \, dx \). Here, \( v = \ln x \).

Key Concepts

Understanding Integration Techniques

Integration techniques are various methods used to find the antiderivative, or integral, of functions in calculus. One such powerful technique is integration by parts. It helps in situations where functions are products of other functions, making direct integration challenging. The formula for integration by parts is given by:

• \[ \int u \, dv = uv - \int v \, du \]

In this formula, \( u \) and \( dv \) are parts of the original integrand. You need to select which part to differentiate (\( u \)) and which to integrate (\( dv \)). The success of this technique depends on choosing \( u \) and \( dv \) correctly.
In our example exercise, \( u \) is chosen as \( f(x) \) and \( dv \) as \( \frac{1}{x} dx \). Once the assignments are made, you differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \). This leads us to the antiderivative, making complex integration manageable.

Basics of Calculus

Calculus is a branch of mathematics focused on change and motion. It is composed of two main parts: differentiation and integration. While differentiation explains how to break down or "differentiate" functions to find rates of change, integration is its counterpart. It combines parts to calculate area or "integrate" functions to find antiderivatives or total accumulations.
Integration by parts, as discussed, is a part of integral calculus. It builds on the basic idea of reversing differentiation through a systematic approach. It allows for calculation of the antiderivative of complex functions.
Calculus is fundamental to science and engineering, allowing the description and prediction of dynamic systems. From determining velocities to calculating area under curves, calculus is essential. Understanding these foundational concepts equips students to tackle more complex mathematical challenges with confidence.

Exploring Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that represent the reverse of differentiation. If you have a function \( f(x) \), its antiderivative F(x) is a function whose derivative is the original function, that is, \( F'(x) = f(x) \).

• Consider the basic function \( f(x) = x \). Its antiderivative would be \( F(x) = \frac{x^2}{2} + C \), where \( C \) stands for the constant of integration.
• For more complex functions, techniques like integration by parts help find these antiderivatives.

In the original exercise provided, \( v = \ln x \) is discovered by integrating \( \frac{1}{x} \), giving us a glimpse into how a seemingly simple function integrates into a logarithmic expression.
Finding antiderivatives allows us to calculate areas, volumes, and other quantities described by functions. This concept is crucial in applying mathematics to real-world scenarios where such calculations are necessary.