Problem 53 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{-\ln 4}^{-\ln 2} 2 e^{\theta} \cosh \theta d \theta$$

Step-by-Step Solution

Answer

The value of the integral is \(\frac{3}{32}\).

Step 1: Recognize the integral identity

The integral involves the function \(2 e^{\theta} \cosh \theta\). Note that \(\cosh \theta = \frac{e^\theta + e^{-\theta}}{2}\). We can use this identity to simplify the expression inside the integral.

Step 2: Simplify the integrand

Substitute \(\cosh \theta\) with its definition: \[2e^{\theta}\cosh \theta = 2e^{\theta} \left(\frac{e^{\theta} + e^{-\theta}}{2}\right) = e^{2\theta} + 1.\]Thus, the integral becomes:\[\int_{-\ln 4}^{-\ln 2} \left(e^{2\theta} + 1\right) d\theta.\]

Step 3: Integrate term-by-term

Integrate each term separately:- The integral of \(e^{2\theta}\) is:\[\int e^{2\theta} d\theta = \frac{1}{2}e^{2\theta} + C.\]- The integral of \(1\) is:\[\int 1 \, d\theta = \theta + C.\]

Step 4: Evaluate the definite integral

Using the results from Step 3, evaluate the integral from \(-\ln 4\) to \(-\ln 2\):\[\left[ \frac{1}{2}e^{2\theta} + \theta \right]_{-\ln 4}^{-\ln 2}.\]First, substitute the upper limit, \(-\ln 2\):\[\frac{1}{2}e^{-2\ln 2} - \ln 2 + \ln 2.\]This simplifies to:\[\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}.\]Now substitute the lower limit, \(-\ln 4\):\[\frac{1}{2}e^{-2\ln 4} + \cancel{(+ \ln 4 - \ln 4)} = \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{32}.\]So, the definite integral is:\[\frac{1}{8} - \frac{1}{32} = \frac{4}{32} - \frac{1}{32} = \frac{3}{32}.\]

Step 5: State the final answer

Therefore, the evaluated integral is \(\frac{3}{32}\).

Key Concepts

Hyperbolic Functions

Understanding hyperbolic functions can help unravel complex calculus topics, especially when dealing with integrals. Hyperbolic functions like \( \sinh \theta \) and \( \cosh \theta \) are analogous to the trigonometric functions sine and cosine, but for a hyperbola rather than a circle. \(cosh \theta = \frac{e^{\theta} + e^{-\theta}}{2} \) and \( sinh \theta = \frac{e^{\theta} - e^{-\theta}}{2} \).Here are a few key points about hyperbolic functions:

• They are defined using exponential functions, which makes their properties quite unique.
• These functions are useful in both pure and applied mathematics, especially in fields like engineering and physics.
• Similarity to trigonometric identities helps simplify computations involving hyperbolic expressions.

Learning how to manipulate these functions is crucial to simplifying complex integrals. When you can replace a hyperbolic function with its exponential identity, it often makes integration more straightforward.

Exponential Functions

Exponential functions are core components in calculus due to their natural properties. Exponential functions like \(e^{\theta}\) exhibit continuous growth and appear frequently in modeling real-world phenomena. In our solution, we manipulate these functions for simplification:

• Exponential functions have the unique property whereby the derivative of \(e^{\theta}\) is \(e^{\theta}\), which largely simplifies calculus operations.
• When combined, such as in \(e^{2\theta}\), they can illustrate how continuous compounding works, a concept useful in finance and science.
• This property extends to integration, making exponential function integration a process that mirrors differentiation.

This step involves recognizing these powers and using substitution techniques, leveraging the intrinsic properties of exponential functions.

Integration Techniques

In calculus, integration techniques transform complex expressions into solvable forms. Whether in single-variable mathematics or more advanced studies, these techniques remain fundamental. Here's how they applied to this exercise:

• Recognize and simplify the integrand: This means identifying when parts of the expression can be rewritten or simplified, such as using exponential identities for hyperbolic functions.
• Term-by-term integration: This involves splitting the integral into simpler parts, as shown by integrating \(e^{2\theta}\) and 1 separately, making calculations manageable.
• Definite integrals: These provide a range (from \(-\ln 4\) to \(-\ln 2\) in this case) and require evaluation of the antiderivative at these bounds.

Mastering these techniques is important, as it allows for evaluating integrals more effectively in both theoretical problems and practical applications.