Problem 29 - Applied Calculus

Question

Derive the formula (called a reduction formula): $$ \int x^{n} e^{x} d x=x^{n} e^{x}-n \int x^{n-1} e^{x} d x $$

Step-by-Step Solution

Answer

The reduction formula is \( \int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx \).

Step 1: Understand the Goal

We aim to derive a reduction formula for the integral \( \int x^n e^x \, dx \). This formula will express the integral in terms of another integral with a reduced power of \( x \).

Step 2: Apply Integration by Parts

Utilize the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Choose \( u = x^n \) and \( dv = e^x \, dx \), then differentiate \( u \) and integrate \( dv \): \( du = n x^{n-1} \, dx \) and \( v = e^x \).

Step 3: Substitute into Integration by Parts Formula

Substitute the expressions for \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int x^n e^x \, dx = x^n e^x - \int e^x (n x^{n-1}) \, dx \].

Step 4: Simplify the Expression

Simplify the second integral in the equation by factoring out the constant \( n \): \[ \int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx \].

Step 5: State the Derived Formula

The final reduction formula is derived: \(\int x^n e^x \, dx = x^n e^x - n \int x^{n-1} e^x \, dx \). This allows computing the integral by reducing the power of \( x \).

Key Concepts

Integration by Parts

Integration by Parts is a useful technique in integral calculus, commonly applied when dealing with products of polynomial and exponential functions. The technique is based on the formula: \[ \int u \, dv = uv - \int v \, du \] This method is particularly helpful for integrals involving the product of two different types of functions, like polynomials and exponentials.
To effectively use this technique, we select two functions from the integrand:

• \( u \): the function to be differentiated
• \( dv \): the function to be integrated

Choosing these components wisely is crucial for simplifying the integral. For instance, in the original problem, selecting \( u = x^n \) and \( dv = e^x \, dx \) makes sense as differentiating polynomials is straightforward, and integrating exponential functions is simple due to their unique properties.
Integration by Parts is an invaluable tool that transforms complicated integrals into more manageable ones, often reducing the integral to simpler forms that can be evaluated easily.

Exponential Function

Exponential functions, denoted as \( e^x \), are a pivotal concept in mathematics, especially in calculus. Their unique property is that they remain unchanged when differentiated or integrated.
This characteristic makes them amenable to integration and differentiation, playing an essential role in various calculus problems. In the context of integration by parts, exponential functions are often chosen as the \( dv \) component. This stems from the fact that integrating \( e^x \) yields \( e^x \) again, making the integral straightforward.
Because the exponential function does not develop extra terms upon integration or differentiation, it simplifies the calculation considerably when paired with other functions such as polynomials. Exponential functions exhibit constant growth rates, which is a reason they frequently appear in applications ranging from population models to compound interest calculations. Understanding their behavior is essential for tackling integrals involving products of different function types.

Polynomial Function

Polynomial functions follow a structure of \( a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 \) where each term is a power of \( x \) multiplied by a coefficient.
These functions are prevalent in algebra and calculus due to their straightforward differentiation and integration properties.In integration by parts, often, polynomials are chosen for the \( u \) term because differentiating them results in a decrease in their power, simplifying the expressions considerably. For instance, choosing \( u = x^n \) and differentiating gives \( du = n x^{n-1} \, dx \), illustrating how polynomials reduce neatly when differentiated.
This reduction is advantageous in the derivation of reduction formulas, which express integrals in terms of lower power polynomial integrals. Hence, polynomial functions serve as a key element in breaking complex integrals into simpler parts, building toward solving them step by step.

Integral Calculus

Integral Calculus focuses on the concept of an integral, which represents the area under a curve on a graph, or quantitatively, the accumulation of quantities. Integrals are computed in many ways for various functions ranging from polynomials to trigonometrics and exponentials, each requiring specific techniques.
Reduction formulas, like the one derived in the original exercise, serve as powerful tools in integral calculus. They provide pathways to solve complex integrals by simplifying them into more elementary forms, as seen when integrating products of polynomials and exponentials. Understandably, mastering integral calculus opens the door to solving a wide array of mathematical problems, granting one the ability to determine areas, volumes, and accumulations. Methods like substitution, and integration by parts, alongside reduction formulas, equip learners with versatile strategies to handle diverse integrals, making integral calculus a cornerstone of mathematical analysis.