Problem 53 - University Calculus: Early Transcendentals

Question

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(e^{-x}+4^{x}\right) d x$$

Step-by-Step Solution

Answer

The integral is \( -e^{-x} + \frac{4^{x}}{\ln(4)} + C \).

Step 1: Understanding the Integral Terms

The integral given is \( \int \left(e^{-x} + 4^{x}\right) dx \). This expression consists of two terms: \( e^{-x} \) and \( 4^{x} \). We will integrate each term separately.

Step 2: Integrating the First Term

For the first term \( \int e^{-x} \, dx \), recall the antiderivative of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \). Therefore, the antiderivative of \( e^{-x} \) is \( -e^{-x} \).

Step 3: Integrating the Second Term

The second term \( \int 4^{x} \, dx \) involves an exponential of the form \( a^{x} \). The antiderivative is \( \frac{a^{x}}{\ln(a)} \). Hence, for \( 4^{x} \), the antiderivative is \( \frac{4^{x}}{\ln(4)} \).

Step 4: Combining the Results

Combine the antiderivatives of both terms: \( -e^{-x} + \frac{4^{x}}{\ln(4)} + C \), where \( C \) is the constant of integration.

Step 5: Checking by Differentiation

Differentiate the result \( -e^{-x} + \frac{4^{x}}{\ln(4)} + C \). The derivative \( \frac{d}{dx}[-e^{-x} + \frac{4^{x}}{\ln(4)}] \) is \( e^{-x} + \frac{4^{x} \ln(4)}{\ln(4)} = e^{-x} + 4^{x} \). This confirms our integral because the result matches the original function.

Key Concepts

Antiderivative

The concept of an antiderivative is fundamental in calculus. An antiderivative of a function is essentially a function whose derivative is the original function. This concept is closely linked to the indefinite integral, which represents a family of functions where the derivative of each function in the family is the integrand.

When finding the antiderivative, it's crucial to remember:

• The rules of differentiation, but in reverse.
• The addition of a constant of integration, usually denoted by "C." This constant arises because derivatives of constants are zero, making it impossible to determine the constant from the derivative alone.

In the exercise, the integral of the expression \( e^{-x} + 4^{x} \) was sought. By integrating each term, we found that the antiderivative is \(-e^{-x} + \frac{4^{x}}{\ln(4)} + C\). This sum includes the general constant of integration "C," signifying any constant could fit the solution and make its derivative match the integrand.

Exponential Functions

Exponential functions are a core component of calculus, particularly in integration and differentiation. They are defined as functions of the form \( a^x \), where \( a \) is a positive constant, and \( x \) is a variable. These functions have unique properties, opening up various applications in physics, finance, and other fields.

In this exercise, we've encountered two types of exponential functions: \( e^{-x} \) and \( 4^{x} \). Here are some key points:

• The exponential function \( e^{-x} \) is a natural exponential function, where \( e \) (approximately 2.718) is the base of natural logarithms.
• To find its antiderivative, recall that the process involves reversing the differentiation of \( e^x \), resulting in \(-e^{-x}\).
• The function \( 4^x \) requires more steps, employing the formula \( \frac{a^x}{\ln(a)} \) for integration, where \( \ln \) denotes the natural logarithm.

Understanding these functions' behavior and properties is essential when working with calculus, as they regularly appear in various mathematical contexts and solutions.

Integration by Parts

Integration by parts is a powerful technique derived from the product rule of differentiation. It is particularly useful when dealing with integrals of products of functions, but it also applies when basic integration rules don't easily apply.In the exercise at hand, though not directly used, the strategy exemplifies how integration by parts is generally approached:

• The formula used for integration by parts is \( \int u \ dv = uv - \int v \ du \), where \( u \) and \( dv \) are parts of the original function.
• This method can effectively transform difficult integrals into more manageable forms.
• Although not all integrals explicitly require this method, understanding its application can simplify solving complex integrals.

In exercises that involve several exponential functions, integration by parts might be applied if the function can be split into product components that align well with the integration by parts formula. Thus, knowing when and how to apply this method is an important skill in calculus.