Problem 28 - Calculus: Early Transcendental Functions

Question

Find the general antiderivative. $$\int \frac{\left(e^{x}\right)^{2}-2}{e^{x}} d x$$

Step-by-Step Solution

Answer

The antiderivative of \(\frac{(e^{x})^{2}-2}{e^{x}}\) is \(e^{x} - 2e^{-x} + C\).

Step 1: Simplify the Expression

Start by simplifying the expression under the integral, \(\frac{(e^{x})^{2}-2}{e^{x}}\). This simplifies to \( e^{x} - \frac{2}{e^{x}} \)

Step 2: Separate the Integral

The integral can be separated into two simpler integrals: \( \int e^{x} dx - \int \frac{2}{e^{x}} dx \)

Step 3: Calculate the First Integral

The antiderivative of \( e^{x} \) is \( e^{x} \). By the integral rule, the result is \( e^{x} \)

Step 4: Calculate the Second Integral

The function \(\frac{2}{e^{x}}\) can also be written as \(2e^{-x}\). The antiderivative of \(2e^{-x}\) is \(-2e^{-x}\). Thus the result of the second integral is \(-2e^{-x}\)

Step 5: Combine the Two Parts

By adding the results of the two integrals and adding a constant 'C' at the end (as the antiderivative introduces an unknown constant), the final result is \(e^{x} - 2e^{-x} + C\)

Key Concepts

Integrals

Integrals are a fundamental concept in calculus, representing the accumulation of quantities. By finding an antiderivative, we solve the integral of a function. In simple terms, an integral will help you find areas beneath curves or accumulate values over an interval. When you encounter an integral, like \(\int f(x) \, dx\), you’re looking for a function, \(F(x)\), whose derivative is \(f(x)\). This function \(F(x)\) is known as an antiderivative.
In this exercise, we are given a more complex expression \(\int \frac{(e^{x})^{2}-2}{e^{x}} \, dx\). The key to solving integrals like this is to first simplify the expression if possible, which makes the integral much easier to solve.

This process involves:

• Recognizing that to integrate requires breaking up complex rational functions into simpler parts.
• Separation of the integral into manageable, smaller integrals.
• Addition of all solved integrals and including a constant of integration \( C \) to capture the general form of the antiderivative.

This helps us get a complete picture of what the antiderivative would look like.

Exponential Functions

Exponential functions are a type of function where the variable appears as an exponent. In mathematical notation, this function has the form \(f(x) = a^{x}\), where \(a\) is a constant. In calculus, exponential functions often involve the number \(e\), known as Euler’s number, which is approximately 2.718.
Understanding these functions is crucial, as they have unique properties that affect their integrals. Specifically:

• The derivative of an exponential function \(e^{x}\) is \(e^{x}\), which makes integrating it straightforward.
• When dealing with negative exponents such as \(e^{-x}\), remember that the integral yields \(-e^{-x}\).

Knowing these properties, you can better handle problems involving exponential functions. In our exercise, recognizing that \(\frac{2}{e^{x}}\) can be rewritten as \(2e^{-x}\) made our task easier, as the integral for \(2e^{-x}\) is simply found using the standard rules.
This understanding streamlines the solution of integrals involving exponential functions.

Simplifying Expressions

Simplifying expressions underpins successful integration. It involves reducing complexity to make mathematical operations easier and more intuitive. When you simplify, you aim to express information in its most compact form. This usually involves:

• Combining like terms.
• Canceling out parts of fractions if possible.
• Recognizing and applying rules of arithmetic and algebra effectively.

In our problem, the original expression \(\frac{(e^{x})^{2}-2}{e^{x}}\) needed simplification before we could tackle integrating it. By simplifying it to \(e^{x} - \frac{2}{e^{x}}\), it allowed us to clearly see two separate integrals that are much simpler to compute.

The simplification helps in understanding structures and forms within expressions, leading to quicker, more efficient integration. Always look for ways to simplify as a first step when you approach integrals. It transforms the problem into something much more approachable, setting a solid foundation for further calculus operations.