Problem 25 - Applied Calculus

Question

Find the exact area. $$\text { Between } y=\ln x \text { and } y=\ln \left(x^{2}\right) \text { for } 1 \leq x \leq 2$$

Step-by-Step Solution

Answer

The exact area is \( \ln(4) - 1 \).

Step 1: Understand the curves

The curves given are \( y = \ln(x) \) and \( y = \ln(x^2) \). We know that \( y = \ln(x^2) = 2\ln(x) \). These represent natural logarithmic functions where \( y = \ln(x) \) is a standard logarithmic curve and \( y = 2\ln(x) \) is a vertically stretched version.

Step 2: Find intersection points

To find where these curves intersect within the interval \([1, 2]\), we equate them: \( \ln(x) = 2\ln(x) \), which simplifies to \( \ln(x) = \ln(x^2) \). This implies \( x = 1 \) and \( x = 2 \), confirming the provided bounds.

Step 3: Determine upper and lower curves

Over the interval \([1, 2]\), the curve \( y = 2\ln(x) \) is above \( y = \ln(x) \). This is seen by noting that \( 2\ln(x) - \ln(x) = \ln(x) > 0 \) for \( x > 1 \). Hence, \( y = 2\ln(x) \) is the upper curve, and \( y = \ln(x) \) is the lower curve between 1 and 2.

Step 4: Set up the integral for area

The area between two curves from \( x = a \) to \( x = b \) is calculated by integrating the difference of the functions: \( \int_a^b [f(x) - g(x)] \, dx \). Here, \( f(x) = 2\ln(x) \) and \( g(x) = \ln(x) \), so the expression becomes \( \int_1^2 [2\ln(x) - \ln(x)] \, dx = \int_1^2 \ln(x) \, dx \).

Step 5: Integrate the resulting function

The integral \( \int \ln(x) \, dx \) can be solved using integration by parts, where \( u = \ln(x) \) and \( dv = dx \), resulting in \( u' = 1/x \), \( v = x \). Thus, \( \int \ln(x) \, dx = x\ln(x) - \int 1 \, dx = x\ln(x) - x + C \).

Step 6: Evaluate the definite integral

Substitute \( x = 2 \) and \( x = 1 \) into the antiderivative to find the definite integral: \[\left[ x\ln(x) - x \right]_1^2 = \left( 2\ln(2) - 2 \right) - \left( 1\ln(1) - 1 \right)\]. Calculate this to find \( (2\ln(2) - 2) - (0 - 1) = 2\ln(2) - 1 \).

Step 7: Simplify the result

The expression \( 2\ln(2) - 1 \) simplifies to \( \ln(4) - 1 \) because \( 2\ln(2) = \ln(2^2) = \ln(4) \). Therefore, the exact area between the curves from \( x = 1 \) to \( x = 2 \) is \( \ln(4) - 1 \).

Key Concepts

Natural logarithm

The natural logarithm, denoted by \( \ln(x) \), is a logarithmic function with the base \( e \), where \( e \) is approximately 2.71828. It is used to determine the power that \( e \) must be raised to obtain the number \( x \).

• The function \( y = \ln(x) \) is defined only for positive values of \( x \), meaning \( x > 0 \).
• It provides a continuous curve that increases as \( x \) increases.
• On a graph, it is essential to note that \( \ln(x) \) passes through the point \( (1,0) \) because \( e^0 = 1 \).

Understanding transformations of the natural logarithm is crucial when analyzing curves. For instance, the function \( y = \ln(x^2) \) demonstrates a property of logarithms where exponentiation inside the logarithm translates to multiplication outside: \( \ln(x^2) = 2\ln(x) \). This represents a stretching of the \( \ln(x) \) curve, doubling its height at any point \( x \). Such transformations help solve problems involving areas between curves, as they reveal relationships between different curve structures.

Integration by parts

Integration by parts is a technique used to solve integrals where standard methods do not suffice. It stems from the product rule of differentiation and helps find the integral of a product of functions. Mathematically, it is expressed as:\[ \int u \, dv = uv - \int v \, du \] Where:

• \( u \) is a function you select to differentiate.
• \( dv \) is a function you select to integrate.
• \( du \) and \( v \) are the derivatives and antiderivatives of \( u \) and \( dv \), respectively.

In the exercise, we used integration by parts to solve \( \int \ln(x) \, dx \). Here, letting \( u = \ln(x) \) and \( dv = dx \) was ideal:

• \( u' = \frac{1}{x} \) and \( v = x \)
• The integral becomes \( x \ln(x) - \int x \cdot \frac{1}{x} \, dx \)
• This reduces to \( x \ln(x) - \int 1 \, dx = x \ln(x) - x + C \)

Integration by parts is a powerful tool in calculus, especially when dealing with functions that involve logarithms and other products.

Definite integral

A definite integral of a function over a closed interval \([a, b]\) represents the net area between the curve of the function and the x-axis. It is calculated as:\[ \int_a^b f(x) \, dx = F(b) - F(a) \] Where \( F(x) \) is the antiderivative of \( f(x) \).

• The definite integral provides a numerical value, as opposed to an antiderivative, which is a general formula.
• In the context of our problem, the definite integral \( \int_1^2 \ln(x) \, dx \) measures the area between the curve \( \ln(x) \) and the curve \( 2\ln(x) \) over the interval from \( x = 1 \) to \( x = 2 \).
• Integrating \( \ln(x) \) involves finding its antiderivative using techniques like integration by parts.

After integrating, substitute the upper and lower limits and subtract to get the exact area. For the exercise, when evaluated between 1 and 2, the result was \( \ln(4) - 1 \). This accurate calculation of area is essential to understanding how different curves interact within given bounds.