Problem 10 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int 2 x^{2} e^{-x} d x $$

Step-by-Step Solution

Answer

\( \int 2 x^{2} e^{-x} \, dx = -(2x^2 + 4x + 4)e^{-x} + C \).

Step 1: Identify Parts for Integration by Parts

The formula for integration by parts is \( \int u \cdot dv = uv - \int v \cdot du \). First, we need to select \( u \) and \( dv \). For the integral \( \int 2x^2 e^{-x} \, dx \), choose \( u = 2x^2 \) and \( dv = e^{-x} \, dx \).

Step 2: Differentiate and Integrate Parts

Calculate \( du \) by differentiating \( u \): \( du = \frac{d}{dx}(2x^2) \, dx = 4x \, dx \). For \( dv = e^{-x} \, dx \), integrate to find \( v \): \( v = \int e^{-x} \, dx = -e^{-x} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the formula: \[ \int 2x^2 e^{-x} \, dx = (2x^2)(-e^{-x}) - \int (-e^{-x})(4x) \, dx \]. Simplify the expression: \[ -2x^2 e^{-x} + 4 \int x e^{-x} \, dx \].

Step 4: Second Integration by Parts

For \( \int x e^{-x} \, dx \), apply integration by parts again. Let \( u = x \) and \( dv = e^{-x} \, dx \). Then \( du = dx \) and \( v = -e^{-x} \).

Step 5: Apply Formula Second Time

Use integration by parts again: \[ \int x e^{-x} \, dx = (-x e^{-x}) - \int (-e^{-x}) \, dx \] which simplifies to \[ -x e^{-x} + \int e^{-x} \, dx = -x e^{-x} - e^{-x} \].

Step 6: Combine Results

Substitute back into the original integration by parts result: \[ -2x^2 e^{-x} + 4(-x e^{-x} - e^{-x}) \]. Expand and simplify: \[ -2x^2 e^{-x} - 4x e^{-x} - 4e^{-x} \].

Step 7: Final Result

Combine all terms to express the final result: \[ -(2x^2 + 4x + 4)e^{-x} + C \], where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in calculus that deals with the accumulation of quantities and the areas under and between curves. An integral, denoted by the symbol \( \int \), is a mathematical object that can be thought of as a generalization of summation over continuous domains. Integrals are used to find areas, volumes, central points, and other quantities that arise when considering infinitesimal data slices.

In the world of calculus, there are primarily two types of integrals: definite and indefinite integrals.

• Definite integrals are computed over a specific interval. This results in a number that represents the total accumulation (area under the curve) from one boundary to another.
• Indefinite integrals, on the other hand, represent a family of functions and include a constant of integration, usually denoted by \( C \).

Integration by parts is a powerful technique derived from the product rule for differentiation. It simplifies the task of finding integrals of products of functions, especially when direct integration is complex. This concept is fundamental when dealing with products of polynomials, exponential functions, and trigonometric functions.

Understanding integral calculus and methods like integration by parts helps solve complex real-world problems involving rates of change and areas beneath curves. With practice, it becomes an invaluable tool for mathematicians, engineers, and scientists.

Exponential Functions

Exponential functions are one of the most important classes of functions in mathematics, characterized by the function form \( f(x) = a^x \) or \( e^x \), where \( e \) is the Euler's number, approximately 2.718. These functions exhibit growth or decay at rates proportional to their current value, which makes them unique and essential in various applications.

Let's break down some key features:

• The base \( e \), known as Euler's number, is the foundation of the natural exponential function \( e^x \). Its unique properties allow it to model continuous growth processes, like population growth or radioactive decay.
• Exponential decay appears in scenarios where quantities decrease rapidly at first and then slow down over time, such as cooling temperatures and diminishing investments.

In calculus, when dealing with exponential functions, the derivative and the integral of \( e^x \) are notably straightforward. For instance, the derivative of \( e^x \) is \( e^x \) itself, and the integral of \( e^x \) also results in \( e^x + C \) (where \( C \) is the constant of integration).

When integrating or differentiating expressions that involve combinations such as products of polynomials and exponentials, integration by parts often comes into play. As seen in the original exercise, selecting the right "u" and "dv" helps unravel these seemingly challenging integrals, making exponential functions a bit more approachable.

Polynomial Integration

Polynomial integration is a straightforward process of integrating expressions involving powers of \( x \), such as \( x^n \). This is a core concept within integral calculus, as polynomials are among the most common functions encountered in mathematics.

The basic rule for integrating a polynomial is given by the power rule for integration:

• For any polynomial term \( x^n \), its integral is \( \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \) and \( C \) is the constant of integration.

In the context of integration by parts, polynomial functions frequently play the role of "\( u \)" because differentiating a polynomial simplifies the function, making calculations easier.

The exercise presented a case of combining polynomials with exponential functions, necessitating the use of integration by parts. Initially setting \( u = 2x^2 \) leverages this simplicity, as differentiating \( u \) brings the exponent down from 2 to 1 (in the form of \( 4x \)), streamlining the process with each integration step.

This integration tactic enhances students' ability to tackle more intricate problems involving combination functions, emphasizing the importance of mastering basic polynomial integration as an essential skill for more advanced calculus.