Problem 39 - Calculus Early Transcendentals

Question

Evaluate each definite integral. \(\int_{0}^{\ln 2} \tanh x d x\)

Step-by-Step Solution

Answer

Question: Evaluate the definite integral of the hyperbolic tangent function from 0 to ln(2). Answer: The definite integral of the hyperbolic tangent function from 0 to ln(2) is equal to ln(5/2).

Step 1: Find the antiderivative of the hyperbolic tangent function

The hyperbolic tangent function is given by the formula \(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\). We can rewrite this as a difference of two fractions: $$\tanh x = \frac{e^x}{e^x + e^{-x}} - \frac{e^{-x}}{e^x + e^{-x}}$$ Now, we can find the antiderivative of the two fractions separately: $$\frac{d}{dx}(\ln(e^{x}+e^{-x}))= \frac{e^x}{e^x + e^{-x}}$$ $$\frac{d}{dx}(-\ln(e^{x}+e^{-x}))= \frac{e^{-x}}{e^x + e^{-x}}$$ Thus, the antiderivative of the hyperbolic tangent function is: $$\int \tanh x dx = \ln(e^{x}+e^{-x}) + C$$

Step 2: Apply the Fundamental Theorem of Calculus

Now, we are ready to apply the Fundamental Theorem of Calculus to evaluate the definite integral. We have: $$\int_{0}^{\ln 2} \tanh x dx = \left[\ln(e^{x}+e^{-x})\right]_0^{\ln 2}$$ Now, we can plug in the limits of integration: $$\left[\ln(e^{x}+e^{-x})\right]_0^{\ln 2} = \left[\ln(e^{\ln 2}+e^{-\ln 2}) - \ln(e^0+e^0)\right]$$ Simplify the expression: $$= \ln\left(\frac{2+ \frac{1}{2}}{1+1}\right) = \ln\left(\frac{5}{2}\right)$$ So, the definite integral of the hyperbolic tangent function from 0 to ln(2) is equal to: $$\int_{0}^{\ln 2}\tanh x dx = \ln\left(\frac{5}{2}\right)$$

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions similar to trigonometric functions, but they are based on hyperbolas rather than circles. They have applications in many fields such as physics, engineering, and calculus. The hyperbolic tangent function, denoted as \(\tanh x\), is one of these functions, defined as:\[tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\]This expression takes the form of a quotient of exponential functions. Each hyperbolic function can be expressed in terms of exponential functions, which helps in deriving their properties and antiderivatives. In this exercise, we're asked to integrate the hyperbolic tangent function \(\tanh x\) over a specific interval. Understanding how to manipulate and simplify these functions is crucial for integration. By rewriting \(\tanh x\) as a difference of two fractions, the integration process becomes manageable. This manipulation leverages the symmetric properties of exponential functions, aiding in the derivation of its antiderivative.

Antiderivative

The antiderivative, often known as the indefinite integral, is a fundamental concept in calculus. It reverses the process of differentiation, allowing us to find a function whose derivative is the given function. For any function \(f(x)\), its antiderivative \(F(x)\) satisfies \(F'(x) = f(x)\).In the context of our exercise, the goal is to find an antiderivative for \(\tanh x\). By decomposing \(\tanh x\) into two simpler fractions, we are able to identify their antiderivatives.

• The first fraction \( \frac{e^x}{e^x + e^{-x}}\) integrates to \(\ln(e^{x} + e^{-x})\).
• The second fraction \( \frac{e^{-x}}{e^x + e^{-x}}\) integrates to \(-\ln(e^{x} + e^{-x})\).

Combining these results gives us the antiderivative:\[\int \tanh x \, dx = \ln(e^{x} + e^{-x}) + C\]where \(C\) is the constant of integration. This means that when differentiating \(\ln(e^{x} + e^{-x})\), we get back \(\tanh x\).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a powerful tool in mathematics. It connects differentiation and integration, two core processes in calculus. The theorem states that if \(F(x)\) is the antiderivative of \(f(x)\), then:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]This means we can evaluate a definite integral by finding an antiderivative and subtracting its values at the bounds of integration.For our problem, once we found the antiderivative of \(\tanh x\) as \(\ln(e^{x} + e^{-x})\), the FTC allows us to compute:\[\int_{0}^{\ln 2} \tanh x \, dx = \left[ \ln(e^{x} + e^{-x}) \right]_0^{\ln 2} = \ln\left(\frac{5}{2}\right)\]By substituting the limits \( 0 \) and \( \ln 2 \), and simplifying, we find the integral's value, demonstrating the utility of FTC in converting the process of integration into straightforward calculations.

Integration Techniques

Integration techniques are methods used to find antiderivatives and evaluate integrals. Important techniques include substitution, integration by parts, and partial fraction decomposition. In working with hyperbolic functions, recognizing forms that directly relate to standard derivatives can simplify integration.In our exercise, the integration technique relies on manipulating \(\tanh x\) into forms that closely resemble expressions with known antiderivatives. We used a method of rewriting \(\tanh x\) as two separate fractions. This step allowed us to derive directly from known derivatives of logarithmic functions:

• Recognizing expressions as derivatives of logarithmic forms simplifies identifying their antiderivatives.
• Breaking down complex functions into sum or differences of simpler forms eases the integration process.

These strategies illustrate how a good grasp of basic derivative and antiderivative forms can aid in solving more complex integration tasks efficiently. Developing intuition for these techniques can significantly simplify solving integrals.