Problem 83 - Applied Calculus

Question

Find an antiderivative and use differentiation to check your answer. $$p(x)=e^{2 x}-e^{-2 x}$$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = \frac{1}{2}e^{2x} + \frac{1}{2}e^{-2x} + C \). Differentiating this gives back \( p(x) = e^{2x} - e^{-2x} \).

Step 1: Identify Antiderivative

Given the function \( p(x) = e^{2x} - e^{-2x} \), we need to find an antiderivative that, when differentiated, returns \( p(x) \). An antiderivative of \( e^{2x} \) is \( \frac{1}{2}e^{2x} \) and an antiderivative of \( e^{-2x} \) is \( -\frac{1}{2}e^{-2x} \). Thus, the antiderivative of \( p(x) \) is \( F(x) = \frac{1}{2}e^{2x} + \frac{1}{2}e^{-2x} + C \), where \( C \) is a constant.

Step 2: Differentiate Antiderivative

To verify the antiderivative, we differentiate \( F(x) = \frac{1}{2}e^{2x} + \frac{1}{2}e^{-2x} + C \). The derivative of \( \frac{1}{2}e^{2x} \) is \( e^{2x} \) and the derivative of \( \frac{1}{2}e^{-2x} \) is \( -e^{-2x} \). The derivative of a constant \( C \) is zero. Therefore, \( F'(x) = e^{2x} - e^{-2x} \), which is \( p(x) \). This confirms the antiderivative is correct.

Step 3: Interpret the Results

The differentiation of \( F(x) \) returned the original function \( p(x) = e^{2x} - e^{-2x} \), confirming that our earlier calculation of the antiderivative was accurate. This confirms that \( \frac{1}{2}e^{2x} + \frac{1}{2}e^{-2x} + C \) is indeed an antiderivative of \( p(x) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes. It is the process used to find a derivative, which represents the slope of a function at any given point. Think of it as a tool to determine how one quantity changes in relation to another.

• The derivative of a function at a particular point gives the slope of the tangent line to the curve at that point.
• Differentiation is a powerful tool for understanding the behavior and characteristics of functions.
• Common differentiation rules include the power rule, product rule, and chain rule. These rules make the process of finding derivatives much easier.

In the context of this exercise, once we found an antiderivative of the original function, we used differentiation to check our work. By taking the derivative of the found antiderivative, we confirmed that it matched the original function. This verification ensures that our antiderivative discovery was correct.

Exponential Functions

Exponential functions are mathematical functions that have the form \( f(x) = a^x \), where \( a \) is a constant and \( x \) is the variable. These functions have unique characteristics that make them important in various fields such as physics, biology, and finance.

• Exponential growth occurs when the base \( a \) is greater than 1, showing an increase over time.
• Conversely, exponential decay happens when \( a \) is between 0 and 1, indicating a decrease.
• Natural exponential functions use the constant \( e \), approximately 2.718, which is the base of natural logarithms. They are written as \( e^x \).

In our exercise, the function \( p(x) = e^{2x} - e^{-2x} \) is composed of natural exponential functions. These special functions' derivatives and integrals follow a pattern due to the nature of \( e \), making them more predictable and often simpler to work with.

Calculus

Calculus is the branch of mathematics focused on studying change and motion. It is divided into two main areas: differentiation and integration. While differentiation looks at rates of change, integration is the process of finding the antiderivative or the area under the curve of a function.

• Calculus explains the behavior of curves and helps in predicting change, which is crucial in sciences and engineering.
• Understanding the relationship between differentiation and integration is central to mastering calculus.
• Both concepts often go hand-in-hand, serving as powerful tools for mathematical modeling and problem-solving.

In the solution to our exercise, calculus principles help us transition from an antiderivative back to the original function via differentiation. This dual nature of calculus—where one operation can verify the other—demonstrates its utility and versatility in mathematics.