Problem 51 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{\ln 2}^{\ln 4} \operatorname{coth} x d x$$

Step-by-Step Solution

Answer

\( \ln(2.5) \)

Step 1: Rewrite the Integral

The integral is given as \( \int_{\ln 2}^{\ln 4} \operatorname{coth} x \ dx \). We remember that \( \operatorname{coth} x = \frac{\cosh x}{\sinh x} \). However, for the purposes of integration, \( \operatorname{coth} x \) can be approached directly using its antiderivative. The antiderivative of \( \operatorname{coth} x \) is \( \ln |\sinh x| \).

Step 2: Apply the Fundamental Theorem of Calculus

The fundamental theorem of calculus tells us that if \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral \( \int_a^b f(x) \ dx \) is \( F(b) - F(a) \). Here, \( F(x) = \ln |\sinh x| \). Evaluating at the bounds, we have:\[\int_{\ln 2}^{\ln 4} \operatorname{coth} x \ dx = \ln |\sinh(\ln 4)| - \ln |\sinh(\ln 2)|\]

Step 3: Simplify the Expression

To simplify the expression, we need to evaluate \( \sinh(\ln a) \) for \( a = 2 \) and \( a = 4 \). Recall that \( \sinh(x) = \frac{e^x - e^{-x}}{2} \), and hence, \( \sinh(\ln a) = \frac{a - \frac{1}{a}}{2} = \frac{a^2 - 1}{2a} \).For \( a = 4 \):\[\sinh(\ln 4) = \frac{4^2 - 1}{2 \times 4} = \frac{15}{8}\]For \( a = 2 \):\[\sinh(\ln 2) = \frac{2^2 - 1}{2 \times 2} = \frac{3}{4}\]

Step 4: Calculate the Result

Now substitute back into the integral expression:\[\ln \left| \frac{15}{8} \right| - \ln \left| \frac{3}{4} \right| = \ln \left( \frac{15}{8} \div \frac{3}{4} \right) = \ln \left( \frac{15}{8} \times \frac{4}{3} \right)\]Simplify this:\[= \ln \left( \frac{60}{24} \right) = \ln(2.5)\]

Step 5: Final Answer

The evaluated integral is:\[\ln(2.5)\]

Key Concepts

Integral Calculals

Integral calculus is the study of integrals and their properties. Essentially, it involves finding the total value, area, or accumulated quantity described by a function. The integral, denoted generally by \( \int \, f(x) \, dx \), represents the accumulation of quantities whose rate of change is given by the function \( f(x) \).

When integrating, there are two main types to consider:

• Indefinite Integrals: This does not have specific limits of integration and results in a family of functions (antiderivatives).
• Definite Integrals: This has specified limits of integration, providing a numerical value that represents accumulation between those bounds.

The given exercise involves a definite integral, specifically from \( \ln 2 \) to \( \ln 4 \). Our task was to evaluate this integral using the function \( \operatorname{coth} x \). Integrals are crucial in applications across physics, engineering, and economics to calculate things like areas under curves, distances, and quantities of materials.

Antiderivatives

Antiderivatives, often called indefinite integrals, are the reverse process of differentiation. In simpler terms, if \( F(x) \) is an antiderivative of \( f(x) \), then differentiating \( F(x) \) yields \( f(x) \). This means that \( F'(x) = f(x) \).

To solve the original exercise, we explored finding the antiderivative of \( \operatorname{coth} x \). Here, the antiderivative of \( \operatorname{coth} x \) is \( \ln |\sinh x| \). Understanding antiderivatives is crucial because they enable us to compute definite integrals using specified limits.

Antiderivatives also have a constant \( C \) of integration when not evaluated between limits, illustrating there can be many functions that derive to the same function. This constant becomes essential when dealing with boundary value problems or initial conditions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration, showing that integration is, in essence, the opposite of differentiation. It consists of two parts:

• Part 1: Establishes that if \( f(x) \) is a continuous function on an interval, then the function defined as the integral is an antiderivative.
• Part 2: Confirms that if \( F(x) \) is any antiderivative of \( f(x) \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).

In our exercise, we applied the second part of the theorem. This allowed us to find the definite integral of \( \operatorname{coth} x \) by evaluating the antiderivative \( F(x) = \ln |\sinh x| \) at the upper limit \( \ln 4 \) and subtracting the result of evaluating it at the lower limit \( \ln 2 \). This process simplifies finding the exact accumulated difference, providing us with the answer \( \ln(2.5) \).

This theorem is fundamental in understanding calculus because it confirms that understanding derivatives allows us to solve integrals, thereby simplifying complex real-world problems across different domains.