Problem 337 - Calculus Volume 1

Question

Find the area between ln x and the x-axis from x = 1 to x = 2.

Step-by-Step Solution

Answer

The area is \( 2 \ln 2 - 1 \).

Step 1: Set up the Integral

To find the area between the curve \( y = \ln x \) and the x-axis from \( x = 1 \) to \( x = 2 \), we need to calculate the definite integral of \( \ln x \) from 1 to 2. The integral is set up as follows: \[ \int_{1}^{2} \ln x \, dx. \]

Step 2: Perform Integration by Parts

Integration by parts is useful for this integral. Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). Let \( u = \ln x \) and \( dv = dx \). Then, \( du = \frac{1}{x} \, dx \) and \( v = x \). Substituting these in gives: \[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx. \]

Step 3: Simplify the Expression

The integral \( \int 1 \, dx = x \), so substituting gives \( x \ln x - x \). Therefore, \( \int \ln x \, dx = x(\ln x - 1) \).

Step 4: Evaluate the Definite Integral

Now, evaluate the definite integral from 1 to 2: \[ \left[x(\ln x - 1)\right]_{1}^{2} = \left[2(\ln 2 - 1)\right] - \left[1(\ln 1 - 1)\right]. \] Since \( \ln 1 = 0 \), this simplifies to \[ 2(\ln 2 - 1) + 1. \]

Step 5: Simplify the Result

Simplify the expression \( 2(\ln 2 - 1) + 1 \) to find: \[ 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1. \] This represents the area between the curve \( y = \ln x \) and the x-axis from \( x = 1 \) to \( x = 2 \).

Key Concepts

Integration by Parts

Integration by parts is a technique used to simplify complex integrals. It is particularly useful when dealing with the product of two functions. The method is based on the formula \( \int u \, dv = uv - \int v \, du \). Here’s how it works:

• Choose one function to be \( u \) (usually the one that simplifies when differentiated).
• Set the other function as \( dv \) (which will be integrated).

In the context of finding the area under \( y = \ln x \), we set \( u = \ln x \), resulting in \( du = \frac{1}{x} \, dx \). We choose \( dv = dx \), which means \( v = x \). Substituting into the formula yields: \( x \ln x - \int x \cdot \frac{1}{x} \, dx \). Further simplification is required by evaluating \( \int x \cdot \frac{1}{x} \, dx = \int 1 \, dx \), which results in \( x \). The final simplified form is essential for calculating the definite integral.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is the logarithm to the base \( e \) (where \( e \approx 2.718 \)). It possesses unique properties that make it a fundamental mathematical function for various fields.

• Derivative: The derivative of \( \ln x \) is \( \frac{1}{x} \), which is crucial in calculus.
• Logarithm of One: \( \ln 1 = 0 \), simplifying many equations and integrals.
• Monotonicity: \( \ln x \) is increasing for \( x > 0 \), influencing the area calculations.

For solving the given exercise, these properties allow us to utilize integration by parts effectively, breaking down the integral of \( \ln x \) for a clearer solution.

Area Under the Curve

Calculating the area under a curve is a common application of definite integrals. In this case, it involves determining the space between the function \( y = \ln x \) and the x-axis from \( x = 1 \) to \( x = 2 \). The definite integral is set up as:\[\int_{1}^{2} \ln x \, dx\]

• Definite Integral Evaluation: Evaluating this integral involves substitution and simplification, especially after using integration by parts.
• Simplification: After integration, plug in the boundary values to get the exact numerical area.

In the solution, we find that the expression resolves to \( 2 \ln 2 - 1 \), which is the exact area between the curve and the x-axis within the specified limits. Understanding this process helps appreciate how calculus allows us to calculate real-world quantities like areas.