Problem 26 - Calculus: Early Transcendentals

Question

Set up the integral to compute the arc length of the function on the given interval. Do not evaluate the integral. $$ f(x)=\frac{1}{x} \text { on }[1,2] $$

Step-by-Step Solution

Answer

\[ L = \int_{1}^{2} \sqrt{1 + \frac{1}{x^4}} \, dx \]

Step 1: Identify the Arc Length Formula

The arc length of a curve given by a function \( f(x) \) from \( x = a \) to \( x = b \) is calculated using the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx \]For \( f(x) = \frac{1}{x} \), we will use this formula to set up the integral.

Step 2: Differentiate the Function

To find the derivative of the function \( f(x) = \frac{1}{x} \), use the power rule: \[ f'(x) = -\frac{1}{x^2} \]

Step 3: Set Up the Integrand

Substitute \( f'(x) = -\frac{1}{x^2} \) into the integrand of the arc length formula: \[ \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} = \sqrt{1 + \frac{1}{x^4}} \]

Step 4: Set Up the Integral

Substitute the integrand \( \sqrt{1 + \frac{1}{x^4}} \) into the integral formula over the interval \([1, 2]\): \[ L = \int_{1}^{2} \sqrt{1 + \frac{1}{x^4}} \, dx \]This integral represents the arc length of the function on the specified interval.

Key Concepts

Integral Setup

To compute the arc length of a function, the first step involves setting up an integral that represents the arc length formula. In this context, "integral setup" means creating the integral from the given function along with its specified interval. For the function \(f(x) = \frac{1}{x}\), our task is to express the arc length over the interval \([1, 2]\) using a specific formula.
The arc length formula is crucial for this step. It is expressed as:

• \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx \)

This formula integrates the square root of \(1 + \left( \frac{d}{dx} f(x) \right)^2\) over the interval from \(x = a\) to \(x = b\), setting up the function \(f(x)\) in the process. Here, \(a = 1\) and \(b = 2\). Thus, effective integral setup requires precision in representing both the function and its bounds correctly.

Differentiation

Differentiation is a key step in determining the expression under the square root in the arc length formula. It involves finding the derivative of the function, which will then be squared and added to the number 1 in the integrand. For the function \(f(x) = \frac{1}{x}\), differentiation is approached through the power rule.
The power rule states: If \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\). In the case of \(\frac{1}{x}\), it can be rewritten as \(x^{-1}\). Differentiating yields:

• \(f'(x) = -x^{-2} = -\frac{1}{x^2}\)

Understanding this derivative is vital, as it defines the additional complexities in the integrand for the arc length calculation. Differentiation transforms a relatively simple function into a more intricate expression, essential for the integrand.

Arc Length Formula

The arc length of a function over a specific interval is calculated using the arc length formula. This formula requires substituting the derivative found in the differentiation step into a pre-defined equation. The arc length formula arithmetically depends on the following:

• \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx \)

To apply this to \(f(x) = \frac{1}{x}\), substitute \(f'(x) = -\frac{1}{x^2}\) into the formula:

• \(\sqrt{1 + \left(-\frac{1}{x^2}\right)^2} = \sqrt{1 + \frac{1}{x^4}}\)

This structured method represents how the formula seamlessly incorporates the derivative into a workable integral setup. Fully understanding this application clarifies the role of each component, facilitating the transition from a basic function to the integral of its arc length.

Definite Integral

In the context of arc length, a definite integral determines the total length of the curve from one point to another. It integrates the derivative-based expression over a closed interval. For the given interval \([1, 2]\), we substitute the previously discussed integrand into the definite integral form:
The integral for arc length here becomes:

• \( L = \int_{1}^{2} \sqrt{1 + \frac{1}{x^4}} \, dx \)

A definite integral like this is bounded and represents the sum of infinitely small segments along the curve between \(x = 1\) and \(x = 2\). While evaluating the integral provides the exact arc length of the curve, setting it up correctly is the goal here. This ensures that the integral accurately captures the geometric properties of the curve across the specified interval.