Problem 336 - Calculus Volume 1

Question

[T] Find the arc length of ln x from x = 1 to x = 2.

Step-by-Step Solution

Answer

The arc length is approximately 2.784.

Step 1: Understand the Formula for Arc Length

The formula to calculate the arc length of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by:\[L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]For this exercise, the function is \( y = \ln x \).

Step 2: Calculate the Derivative

First, find the derivative of \( y = \ln x \). The derivative \( \frac{dy}{dx} \) is:\[\frac{d}{dx}(\ln x) = \frac{1}{x}\]

Step 3: Plug Into the Arc Length Formula

Substitute the derivative into the arc length formula:\[L = \int_1^2 \sqrt{1 + \left( \frac{1}{x} \right)^2} \, dx\]Simplified, it becomes:\[L = \int_1^2 \sqrt{1 + \frac{1}{x^2}} \, dx\]

Step 4: Simplify the Integrand

Simplify the expression under the square root:\[L = \int_1^2 \sqrt{\frac{x^2 + 1}{x^2}} \, dx = \int_1^2 \frac{\sqrt{x^2 + 1}}{x} \, dx\]

Step 5: Change of Variables

Make a substitution to simplify the integral further. Let \( u = x^2 + 1 \), then \( du = 2x \, dx \), or \( x \, dx = \frac{du}{2} \). Change limits: when \( x = 1, u = 2 \); when \( x = 2, u = 5 \).\[L = \int_2^5 \frac{\sqrt{u}}{2} \, du\]

Step 6: Integrate

Calculate the integral:\[L = \frac{1}{2} \int_2^5 u^{1/2} \, du\]The integral of \( u^{1/2} \) is \( \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \). Thus, the integral becomes:\[L = \frac{1}{2} \times \frac{2}{3} [u^{3/2}]_2^5 = \frac{1}{3} [u^{3/2}]_2^5\]

Step 7: Evaluate the Definite Integral

Substitute the limits of the integral:\[L = \frac{1}{3} (5^{3/2} - 2^{3/2})\]Calculate: \( 5^{3/2} = \sqrt{5^3} = \sqrt{125} \approx 11.18 \), \( 2^{3/2} = \sqrt{2^3} = \sqrt{8} \approx 2.828 \).Then:\[L = \frac{1}{3} (11.18 - 2.828) \approx \frac{1}{3} \times 8.352 \approx 2.784\]

Step 8: Conclusion

The arc length of \( \ln x \) from \( x = 1 \) to \( x = 2 \) is approximately \( 2.784 \).

Key Concepts

Integration

Integration is a fundamental concept in calculus, used to calculate areas under curves, volumes, and lengths, such as arc lengths. In this exercise, integration helps find the arc length of the curve described by the natural logarithm function, from one point to another along the x-axis.

When working with arc lengths, we set up the integral with the special formula:

• For a function \( y = f(x) \) from \( x = a \) to \( x = b \), the arc length \( L \) is given by \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

First, we calculate the derivative \( \frac{dy}{dx} \) and substitute it into the formula. The integral often requires algebraic manipulation to simplify it so it can be solved more easily.

Understanding integration and its application in these kinds of problems is crucial, as it aids in calculating various physical lengths and measurements, extending beyond theoretical mathematics.

Natural Logarithm

The natural logarithm is a specific kind of logarithm, where the base is the mathematical constant \( e \), approximately 2.718. It is symbolized by \( \ln(x) \) and is widely used for its unique properties in calculus and mathematical modeling.

For the function \( y = \ln(x) \), the derivative is \( \frac{1}{x} \). This derivative is a fundamental element in determining how the function \( \ln(x) \) changes. When understanding the arc length problem, remember the properties of the natural_logarithm:

• It increases as \( x \) increases while always being positive.
• The derivative \( \frac{dy}{dx} = \frac{1}{x} \) means the slope decreases as \( x \) increases.

The behavior of \( \ln(x) \) makes integration straightforward once we adjust the integrand formula with this derivative.

Substitution Method

The substitution method in integration is a critical technique that simplifies complex integrals by changing variables.

This exercise required simplifying the arc length formula integrated form of \( \int \frac{\sqrt{x^2 + 1}}{x} \, dx \). The substitution method makes this easier by choosing a new variable.

• In this case, let \( u = x^2 + 1 \), allowing the derivative \( du = 2x \, dx \), which simplifies to \( x \, dx = \frac{du}{2} \).

This substitution transforms the integral into a more straightforward \[ L = \int_2^5 \frac{\sqrt{u}}{2} \, du \], which is much easier to handle.

Substitution is powerful because it turns complex problems into ones that can be solved using basic integral rules, often leading to more elegant solutions that save effort and prevent errors.