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Assume that a is a positive constant. Find the general antiderivative of the given function. $$ f(x)=\frac{e^{-a x}+e^{a x}}{2 a} $$

Short Answer

Expert verified
The general antiderivative is \( \frac{1}{2a^2} (e^{ax} - e^{-ax}) + C \).

Step by step solution

01

Simplify the Function

The given function is \( f(x) = \frac{e^{-a x} + e^{a x}}{2a} \). Let us simplify this expression to prepare for antiderivation. We can separate the fractions as \( \frac{e^{-ax}}{2a} + \frac{e^{ax}}{2a} \).
02

Identify the Basic Antiderivatives

Recall that the antiderivative of \( e^{kx} \) with respect to x is \( \frac{1}{k}e^{kx} + C \), where \( k \) is a constant and \( C \) is the integration constant. We apply this knowledge to each term separately.
03

Antiderivation of the First Term

For \( \frac{e^{-ax}}{2a} \), we use the antiderivative formula: the antiderivative is \( \frac{1}{-a} e^{-ax} \). Thus, this part becomes \( \frac{1}{2a} \cdot \frac{1}{-a} e^{-ax} = -\frac{1}{2a^2} e^{-ax} \).
04

Antiderivation of the Second Term

For \( \frac{e^{ax}}{2a} \), again use the formula. The antiderivative is \( \frac{1}{a} e^{ax} \), hence this term becomes \( \frac{1}{2a} \cdot \frac{1}{a} e^{ax} = \frac{1}{2a^2} e^{ax} \).
05

Combine the Antiderivatives

Add the results of Steps 3 and 4: \( -\frac{1}{2a^2} e^{-ax} + \frac{1}{2a^2} e^{ax} + C \), where \( C \) is the integration constant. Simplifying this yields the general antiderivative: \( \frac{1}{2a^2} (e^{ax} - e^{-ax}) + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are fundamental in calculus and many applications in science and engineering. An exponential function is generally of the form \( e^{kx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718. The constant \( k \) determines the rate of growth or decay. For \( k > 0 \), the function represents exponential growth, while \( k < 0 \) implies exponential decay.

In the exercise, we encountered both \( e^{-ax} \) and \( e^{ax} \). Each term shows how exponential functions behave differently by merely changing the sign of the constant \( a \). Understanding these functions' behaviors helps in anticipating the solution of their antiderivatives in calculus. This understanding is essential as it provides insight into the symmetry and transformations of these functions when integrating.
  • Exponential growth/decay functions show changes in quantity over time.
  • These functions are continuously increasing or decreasing.
  • Their antiderivatives are closely related to their original expressions.
Integration Constant
The integration constant, often denoted as \( C \), plays a key role in indefinite integration. Whenever you find an antiderivative, remember that it’s there to represent the family of all possible curves that differ only by a vertical shift. This constant can take any real number value and makes sure your antiderivative is general.

This role of \( C \) ensures that each function's antiderivative is unique up to an additive constant. Therefore, each member of the family of functions has the same derivative, highlighting the beautiful symmetry in calculus. From techniques of finding antiderivatives to solving real-world applications, accounting for the integration constant is crucial.
  • Always add \( C \) when computing indefinite integrals.
  • The constant ensures the generality and completeness of solutions.
  • It can be determined if initial conditions are given, making solutions specific.
Simplification in Calculus
Simplification is a powerful tool in calculus that makes solving problems much more manageable. Before tackling antiderivatives, a common strategy is to simplify the function if possible. Simplification involves rewriting complex expressions in a more straightforward form. This approach often involves breaking down functions, much like we did by separating the original function into two simpler fractions that could each be individually integrated.

By simplifying, we reduce the likelihood of error during integration and make the overall process smoother. This step can often involve algebraic manipulation, like separating terms or factoring, to identify patterns or use known antiderivatives directly.
  • Clarifies the path to the solution by reducing complexity.
  • Eliminates unnecessary steps, focusing directly on integration.
  • Reveals potential use of calculus rules, such as linearity and substitution.

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