Problem 34 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int\left(2 e^{x}-3 e^{-2 x}\right) d x\end{equation}

Step-by-Step Solution

Answer

\( \int (2e^x - 3e^{-2x}) \, dx = 2e^x + \frac{3}{2}e^{-2x} + C \).

Step 1: Identify the Components

The given function to integrate is \( 2e^x - 3e^{-2x} \). We need to evaluate its integral with respect to \( x \). The integral can be rewritten as \( \int (2e^x) \, dx - \int (3e^{-2x}) \, dx \).

Step 2: Integrate \(2e^x\)

The first term, \( 2e^x \), is a simple exponential function. The integral of \( e^x \) with respect to \( x \) is \( e^x \). Therefore, the integral of \( 2e^x \) is \( 2e^x + C_1 \), where \( C_1 \) is a constant of integration.

Step 3: Integrate \(-3e^{-2x}\)

The second term is \( -3e^{-2x} \). To integrate this, notice the integral of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \), assuming \( a eq 0 \). Thus, the integral of \( -3e^{-2x} \) is \( -3 \cdot \left(-\frac{1}{2}e^{-2x}\right) = \frac{3}{2}e^{-2x} + C_2 \), where \( C_2 \) is a constant of integration.

Step 4: Combine the Results

Combining the results from Steps 2 and 3, we have \( 2e^x + \frac{3}{2}e^{-2x} + C \), where \( C = C_1 + C_2 \) represents the overall constant of integration for the indefinite integral.

Key Concepts

Exponential Functions

Exponential functions are a core topic in calculus and play a vital role, especially in integration. An exponential function is of the form \( f(x) = ae^{bx} \), where \( a \) and \( b \) are constants, and \( e \) is approximately equal to 2.718. This special number, known as Euler's number, is the base of natural logarithms. Exponential functions are unique because they have the same rate of change as their actual value. This means the derivative or slope of \( e^x \) is also \( e^x \).

One property of exponential functions, which simplifies integration, is that they can model growth or decay. For instance, \( e^x \) represents continuous growth, whereas \( e^{-x} \) represents continuous decay. This trait makes exponential functions easy to integrate, especially since the integral of \( e^x \) is consistently \( e^x \) again, plus a constant. This simplicity extends even to functions with a coefficient, as in \( 2e^x \) or \( -3e^{-2x} \), as these coefficients can be directly factored out without complicating the integration process.

Definite and Indefinite Integrals

Integrals are fundamental in calculus, helping us find the area under curves. There are two main types of integrals: definite and indefinite integrals.

• Indefinite Integrals: These are integrals without upper or lower limits, providing an antiderivative or a family of functions. The result always includes a constant of integration, typically denoted as \( C \). For example, the indefinite integral of \(2e^x\) gives \(2e^x + C\).
• Definite Integrals: In contrast, these involve limits and yield a numerical value representing the area under the curve between two points. The constant \( C \) cancels out when computing definite integrals because the limits provide specific values.

Indefinite integrals are crucial for basic integration skills and understanding the principles of calculus. They form the basis upon which you can build your skills for solving more complex problems involving definite integrals, which often require evaluating the constant within specific bounds.

In the context of our original exercise, we focused on finding indefinite integrals, integrating each term separately, and then combining them, including the constant \( C \) to represent any antiderivative.

Integration Techniques

Integration techniques are essential tools for solving complex integrals. When facing integrals, understanding the appropriate strategy is vital for evaluating quickly and accurately.

• Basic Integration: This is the simplest method, often applied when you deal with straightforward functions such as \(e^x\). You directly use fundamental integration rules, making this technique very efficient.
• Integration by Parts: This technique is useful for products of functions and relies on the formula \( \int u \, dv = uv - \int v \, du \). It’s particularly handy for polynomials or when basic methods don’t suffice.
• Substitution Method: Often known as \( u \)-substitution, this method simplifies complex integrals. By substituting part of the integral with a single variable, you turn it into a basic form that is easy to integrate.

In our original exercise, the problem yielded to straightforward integration, exploiting the predictability of exponential functions: the integral of \(e^{ax}\) is always \(\frac{1}{a}e^{ax}\), adjusted by any constants included in the function. Having a grasp of these basic techniques helps in quickly identifying the simplest method for solving integrals efficiently.

With practice, choosing the right technique becomes intuitive, allowing for smooth problem-solving of both straightforward and complex integrals.