Problem 67 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(67-74\) in terms of a. inverse hyperbolic functions. b. natural logarithms. $$\int_{0}^{2 \sqrt{3}} \frac{d x}{\sqrt{4+x^{2}}}$$

Step-by-Step Solution

Answer

\( \ln(\sqrt{3} + 2) \)

Step 1: Identify the Integral Form

The integral \( \int \frac{dx}{\sqrt{a^2 + x^2}} \) is a standard form that can be solved using inverse hyperbolic functions. It is also a variation of the integral that can be evaluated in terms of natural logarithms.

Step 2: Solve in terms of Inverse Hyperbolic Functions

The integral \( \int \frac{dx}{\sqrt{a^2 + x^2}} = \text{sinh}^{-1}\left(\frac{x}{a}\right) + C \) can be applied directly. For \( a = 2 \), the integral becomes:\[\int \frac{dx}{\sqrt{4 + x^2}} = \text{sinh}^{-1}\left(\frac{x}{2}\right) + C\]Evaluate from 0 to \( 2\sqrt{3} \):\[\left[ \text{sinh}^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2\sqrt{3}} = \text{sinh}^{-1}\left(\frac{2\sqrt{3}}{2}\right) - \text{sinh}^{-1}\left(0\right)\]\[= \text{sinh}^{-1}(\sqrt{3})\]Since \( \text{sinh}^{-1}(\sqrt{3}) = \ln(\sqrt{3} + 2) \), the result is \( \ln(\sqrt{3} + 2) \).

Step 3: Solve in terms of Natural Logarithms

The equivalent integral in terms of logarithms is \( \int \frac{dx}{\sqrt{a^2 + x^2}} = \ln|x + \sqrt{a^2 + x^2}| + C \). For \( a = 2 \):\[\int \frac{dx}{\sqrt{4 + x^2}} = \ln|x + \sqrt{4 + x^2}| + C\]Evaluate from 0 to \( 2 \sqrt{3} \):\[\left[ \ln(x + \sqrt{4 + x^2}) \right]_0^{2\sqrt{3}} = \ln(2\sqrt{3} + \sqrt{4 + 12}) - \ln(0 + 2)\]\[= \ln(2\sqrt{3} + 4) - \ln(2)\]\[= \ln\left(\frac{2\sqrt{3} + 4}{2}\right) = \ln(\sqrt{3} + 2)\]

Step 4: Conclude the Calculation

Both methodologies provide the same result. Thus, the evaluation of the integral \( \int_{0}^{2 \sqrt{3}} \frac{d x}{\sqrt{4+x^{2}}} \) in terms of both inverse hyperbolic functions and natural logarithms is \( \ln(\sqrt{3} + 2) \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on the concept of integration. The primary goal of integral calculus is to find the total accumulation or area under a curve. In essence, integration is the reverse of differentiation. It helps in calculating quantities like areas, volumes, and other related measures.
To better understand integration, it's essential to understand the process. When you integrate a function, you essentially sum up infinitely small pieces to calculate a whole quantity. This process is akin to adding slices of bread to make a loaf.

• Definite Integrals: These represent the area under a curve between two specified limits. They provide a number as the result.
• Indefinite Integrals: Also known as antiderivatives, they represent a family of functions and include a constant of integration, often denoted as "C".

The integral addressed in our exercise is a definite integral, where limits are specifically 0 and \(2\sqrt{3}\). This means we are seeking the exact area under the curve of the integrand from these two points.

Natural Logarithms

Natural logarithms are logarithms with a base of \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm, denoted as \(\ln(x)\), often appears in calculus, especially when solving problems related to growth and decay.
In integral calculus, natural logarithms often emerge when evaluating specific integrals, particularly those involving inverse trigonometric and hyperbolic functions. They provide a convenient and powerful tool for simplifying expressions and solving integrals.

• The natural logarithm function has the property \(\ln(ab) = \ln(a) + \ln(b)\), which can be useful in manipulating expressions.
• The derivative of \(\ln(x)\) is \(\frac{1}{x}\), showing the natural logarithm's relationship with exponentials.

In our exercise, evaluating the integral using natural logarithms involved finding an expression in terms of \(\ln\). The solution showed an equivalent result by evaluating forms of functions involving natural logarithms, specifically providing \(\ln(\sqrt{3} + 2)\) as the result.

Definite Integrals

Definite integrals are a way to calculate the exact accumulation of a quantity over an interval. Unlike indefinite integrals, which are expressed as a family of functions with an arbitrary constant, definite integrals provide a concrete numerical result.
The calculation of a definite integral involves finding the antiderivative of a function and then using the limits of integration to compute its actual value. The process can be visualized as the area under a curve over a specific interval.

• The Fundamental Theorem of Calculus connects differentiation and integration, stating that taking the derivative of an antiderivative returns the original function.
• Evaluating a definite integral involves subtracting the value of the antiderivative at the lower limit from its value at the upper limit.

In our specific problem, the integral \(\int_{0}^{2 \sqrt{3}} \frac{dx}{\sqrt{4+x^2}}\) was evaluated over a specific range. Two methods, involving inverse hyperbolic functions and natural logarithms, showed that the integral evaluates to \(\ln(\sqrt{3} + 2)\), demonstrating that definite integrals provide precise, numerical results dependent on the designated interval.