Problem 54 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(51-60 .\) $$ \int_{0}^{\ln 2} 4 e^{-\theta} \sinh \theta d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2 \ln 2 + \frac{3}{4} \).

Step 1: Identify the Problem

We need to evaluate the integral \( \int_{0}^{\ln 2} 4 e^{-\theta} \sinh \theta \, d\theta \). This is a definite integral involving the product of an exponential function and a hyperbolic sine function.

Step 2: Understand the Function Properties

Recall that the hyperbolic sine function, \( \sinh \theta \), can be expressed as \( \frac{e^{\theta} - e^{-\theta}}{2} \). Applying this identity will help simplify the integration.

Step 3: Simplify the Integrand

Substitute \( \sinh \theta = \frac{e^{\theta} - e^{-\theta}}{2} \) into the integral. This gives us: \[ \int_{0}^{\ln 2} 4 e^{-\theta} \left( \frac{e^{\theta} - e^{-\theta}}{2} \right) d\theta = \int_{0}^{\ln 2} 2 (1 - e^{-2\theta}) d\theta. \]

Step 4: Split the Integral

Separate the integral into two simpler integrals: \[ \int_{0}^{\ln 2} 2 (1 - e^{-2\theta}) d\theta = 2 \int_{0}^{\ln 2} d\theta - 2 \int_{0}^{\ln 2} e^{-2\theta} d\theta. \]

Step 5: Integrate Each Part

Evaluate each integral separately. 1. \( 2 \int_{0}^{\ln 2} d\theta = 2[\theta]_{0}^{\ln 2} = 2(\ln 2 - 0) = 2 \ln 2 \).2. \( 2 \int_{0}^{\ln 2} e^{-2\theta} d\theta = 2 \left[ -\frac{1}{2} e^{-2\theta} \right]_{0}^{\ln 2} = -e^{-2\ln 2} + 1 \).

Step 6: Simplify the Expression

For the second part of the integral, \( -e^{-2\ln 2} + 1 = -(\frac{1}{4}) + 1 = \frac{3}{4} \). Combine both parts of the expression: \( 2 \ln 2 + \frac{3}{4} \).

Step 7: Write the Final Answer

The value of the definite integral \( \int_{0}^{\ln 2} 4 e^{-\theta} \sinh \theta \, d\theta \) is \( 2 \ln 2 + \frac{3}{4} \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are important in calculus. They arise naturally in various physical contexts and mathematical equations, such as the solution of the hyperbolic differential equation. One of the key hyperbolic functions is the hyperbolic sine, denoted as \( \sinh \theta \). It is defined as: \[ \sinh \theta = \frac{e^{\theta} - e^{-\theta}}{2} \] This definition shows that \( \sinh \theta \) is closely related to exponential functions. The hyperbolic functions have similar properties to the trigonometric functions but differ in signs. For instance, while the identity \( \sin^2 x + \cos^2 x = 1 \) holds for trigonometric functions, hyperbolic functions satisfy the identity:\( \cosh^2 \theta - \sinh^2 \theta = 1 \). Understanding hyperbolic functions is crucial for solving integrals involving combinations of exponential and hyperbolic terms, as seen in the given problem.

Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent. The most common is the natural exponential function \( e^x \), where \( e \) is approximately 2.718. These functions are pivotal due to their unique properties, such as the derivative and integral of \( e^x \) being \( e^x \). They model growth and decay processes in science, including population growth and radioactive decay. In the integral \( \int 4 e^{-\theta} \sinh \theta \ d\theta \), the exponential function \( e^{-\theta} \) represents a decaying process as \( \theta \) increases. Exponential functions can greatly simplify the integration process when paired with their inverse, which is often a hyperbolic function. Thus, recognizing and manipulating exponential terms are essential in integral calculus, allowing us to simplify and evaluate complex expressions.

Integration Techniques

Integration techniques are methods used to evaluate integrals, particularly useful for complex or non-obvious antiderivatives. Choosing the right technique depends on the function's form. For the integral \( \int_{0}^{\ln 2} 4 e^{-\theta} \sinh \theta \, d\theta \), substitution and simplification by identities help. Initially, express \( \sinh \theta \) using its exponential form: \[ \sinh \theta = \frac{e^{\theta} - e^{-\theta}}{2} \] Substituting this into the original integral simplifies the expression. - **Splitting integrals**: Breaking down a complex integral into parts makes integration manageable, as shown with \[ 2 \int d\theta - 2 \int e^{-2\theta} d\theta. \] - **Evaluating components**: Each part is solved individually, and then combined to find the solution. Integration by substitution and recognizing patterns like exponential decay terms simplifies solving integrals. Familiarity with such techniques is crucial for tackling diverse calculus problems efficiently.