Problem 42 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-62\). $$ \int\left(2 e^{x}-3 e^{-2 x}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2e^x + \frac{3}{2}e^{-2x} + C \).

Step 1: Break Down the Integral

The integral we need to evaluate is \( \int (2e^x - 3e^{-2x}) \, dx \). This integral can be separated into two separate integrals: \( \int 2e^x \, dx \) and \( \int -3e^{-2x} \, dx \).

Step 2: Integrate the First Term

Consider the first integral \( \int 2e^x \, dx \). The integral of \( e^x \) with respect to \( x \) is \( e^x \), so: \[ \int 2e^x \, dx = 2 \int e^x \, dx = 2e^x + C_1 \] where \( C_1 \) is a constant of integration.

Step 3: Integrate the Second Term

Now, consider the second integral \( \int -3e^{-2x} \, dx \). Using the substitution method, let \( u = -2x \), then \( du = -2 \, dx \) implies \( dx = -\frac{du}{2} \). Substitute \( u \) and \( du \): \[ \int -3e^{-2x} \, dx = -3 \int e^u \cdot -\frac{du}{2} = \frac{3}{2} \int e^u \, du = \frac{3}{2} e^u + C_2 \] Substituting back \( u = -2x \): \[ \frac{3}{2} e^{-2x} + C_2 \]

Step 4: Combine Results

Add the results of both integrals from Steps 2 and 3: \[ \int (2e^x - 3e^{-2x}) \, dx = 2e^x + \left(\frac{3}{2}e^{-2x} + C_2\right) = 2e^x + \frac{3}{2}e^{-2x} + C \] where \( C = C_1 + C_2 \) is the combined constant of integration for the whole expression.

Key Concepts

Indefinite Integrals

In the world of calculus, an indefinite integral is an antiderivative. It's like finding the original function when given its derivative. The notation for an indefinite integral is \( \int f(x) \, dx \). Learning the basics helps you understand complex integrals better.
Indefinite integrals include a constant of integration, often denoted by \( C \). This constant emerges because when you differentiate a function, any constant's derivative is zero. So, while integrating, we add this \( C \) to account for any lost constant term.

• It's also important to know that indefinite integrals may result in a family of functions. Any function in this family can have different constants.

Knowing indefinite integrals opens up pathways to solving many real-world problems, especially in physics and engineering, where rates of change are involved.

Exponential Functions

Exponential functions are unique due to their rate of growth. They are defined using the base of natural logarithms, denoted as \( e \), approximately equal to 2.718. These functions, like \( e^x \), have a distinct property where the function is its own derivative.
This means when you integrate \( e^x \), the result is simply \( e^x + C \), showcasing the seamless behavior of exponential functions. They are vital in modeling growth or decay, such as population growth or radioactive decay.

• The exponential function \( e^{-2x} \) involves an extra layer, where \( -2 \) is the coefficient affecting the rate of decay.
• By integrating this function, you gain insight into how changing variables affect the whole expression.

Recognizing these fundamental properties helps in simplifying and solving calculus problems involving exponential terms.

Substitution Method

The substitution method is a powerful tool to simplify integration, especially when dealing with composite functions. It's akin to reversing the chain rule used in differentiation.
When applying this method, you start by identifying a part of the integral to replace with a simpler expression, typically denoted by \( u \).
For example, in the integral \( \int -3e^{-2x} \, dx \), making \( u = -2x \) helps simplify the process.

• Differentiating \( u \) with respect to \( x \), gives you \( du = -2 \, dx \), informing the substitution for \( dx \).
• This transforms the integral into a simpler expression that is easier to evaluate.

The substitution method is especially useful when dealing with exponential or trigonometric integrals in calculus. It greatly reduces the effort needed to find antiderivatives of more complex expressions.