Problem 65 - Advanced Engineering Mathematics

Question

In the study of techniques of integration in calculus, certain indefinite integrals of the form \(\int e^{a x} f(x) d x\) could be evaluated by applying integration by parts twice, recovering the original integral on the right-hand side, solving for the original integral, and obtaining a constant multiple \(k \int e^{a x} f(x) d x\) on the left-hand side. Then the value of the integral is found by dividing by \(k\). Discuss: For what kinds of functions \(f\) does the described procedure work? Your solution should lead to a differential equation. Carefully analyze this equation and solve for \(f\).

Step-by-Step Solution

Answer

The method works for functions \( f(x) \) that are polynomials or exponential functions, i.e., \( f(x) = P(x) + C e^{bx} \), ensuring the recursive form \( I = kI \).

Step 1: Understanding the Problem

We need to find the kind of functions \( f(x) \) where the integral \( \int e^{ax} f(x) \, dx \) can be simplified using integration by parts twice to recover the original integral.

Step 2: Applying Integration by Parts Twice

First, recall that integration by parts is given by \( \int u \, dv = uv - \int v \, du \). To use this method twice, select \( u = f(x) \) and \( dv = e^{ax} dx \). Calculate \( du \) and \( v \), apply formula, then repeat for the resulting integral.

Step 3: Writing the Equation

The goal is to express the integral in terms of itself. After performing integration by parts twice, we obtain an equation of the form \( I = something + k I \). Here, \( I = \int e^{ax} f(x) \, dx \) and \( k \) is a constant.

Step 4: Solving for the Original Integral

Solve the equation \( I = something + kI \) for \( I \). Simplify this to find \( I = \frac{something}{1-k} \). Ensure \( k eq 1 \) to avoid division by zero.

Step 5: Analyzing the Differential Equation

To determine \( f(x) \), recognize that if integration by parts results in \( I = cI \), then it implies a relation for \( f(x) \). Assume an expression for \( f(x) \) like a polynomial or exponential and find \( f(x) \) through substitution.

Step 6: Solve the Differential Equation

Assuming \( f(x) \) is of the form \( f(x) = P(x) + C e^{bx} \), for some polynomial \( P(x) \) and constants \( C, b \), solve for all parameters that satisfy \( I = k I \). Find \( f(x) \) such that the double application of integration by parts returns to \( I \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique in calculus used to solve integrals of products of functions. It is based on the product rule for differentiation and is given by:

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

To successfully apply this method, we need to smartly choose which part of our integral to differentiate (\(u\)) and which to integrate (\(dv\)).
In solving problems like \(\int e^{ax} f(x) \, dx\), start by assigning \(u = f(x)\) and \(dv = e^{ax} \, dx\).Calculate \(du\) by differentiating \(u\), and \(v\) by integrating \(dv\).
Applying integration by parts results in a new integral that often appears different, and repeating the process can sometimes return us to a form similar to the original integral.
This recursiveness allows us to resolve the equation and express our integral in terms of itself, leading to solutions involving a constant multiplier.

Indefinite Integrals

An indefinite integral, often simply called an antiderivative, is a function whose derivative yields the original function being integrated.

• It is represented as \( \int f(x) \, dx = F(x) + C \), where \(F(x)\) is the antiderivative of \(f(x)\), and \(C\) is the constant of integration.

Indefinite integrals provide the general form of solutions rather than a specific numerical result because they include an arbitrary constant.
In the context of \( \int e^{ax} f(x) \, dx \), finding indefinite integrals using integration by parts gives insight into the family of functions \(F(x) + C\).
This family reflects all possible functions that differentiate to produce \(f(x)\), helping solve problems involving functions represented as integrals.
Integral calculus offers tools like integration by parts and substitution to deal with more complex functions, discovering infinitely many antiderivatives for a given function.

Differential Equations

Differential equations are equations that relate a function with its derivatives.

• They are useful in modeling various phenomena such as motion, heat, and population dynamics.

In integration exercises like these, determining the function \(f(x)\) often leads to forming a differential equation.
For example, double application of integration by parts might result in \(I = k I\), implying a relationship between the integral and itself.
This equation can suggest that \(f(x)\) takes a form like \(P(x) + C e^{bx}\), where \(P(x)\) is a polynomial and \(C, b\) are constants.
Solving the differential equation involves substituting \(f(x)\) back into the expression and finding appropriate parameters that satisfy the equation.
This allows us to identify which kinds of functions \(f(x)\) can simplify using integration by parts, leading to deeper understanding of both integration techniques and their connection to differential equation solutions.