Problem 97 - University Calculus: Early Transcendentals

Question

Find the area between the curves \(y=\ln x\) and \(y=\ln 2 x\) from \(x=1\) to \(x=5\).

Step-by-Step Solution

Answer

The area between the curves is \(4 \ln 2\).

Step 1: Identify the functions

The problem involves finding the area between the curves of the functions given by the equations \( y = \ln x \) and \( y = \ln(2x) \) on the interval from \( x = 1 \) to \( x = 5 \).

Step 2: Determine the region of integration

The region of integration is from \( x=1 \) to \( x=5 \), where we need to find where one curve is above the other. Note that \( y=\ln(2x) = \ln x + \ln 2 \), so the function \( y = \ln(2x) \) is always above \( y = \ln x \) in this interval.

Step 3: Set up the integral

The area between the two curves can be found by the integral of the difference of the functions:\[\text{Area} = \int_{1}^{5} [\ln(2x) - \ln x] \, dx.\]

Step 4: Simplify the integrand

Simplify the expression inside the integral: \( \ln(2x) - \ln x = \ln 2 + \ln x - \ln x = \ln 2 \).This simplifies our integral to:\[\text{Area} = \int_{1}^{5} \ln 2 \, dx.\]

Step 5: Evaluate the integral

The integral of a constant \( \ln 2 \) with respect to \( x \) over the interval from 1 to 5 is:\[\int_{1}^{5} \ln 2 \, dx = \ln 2 \cdot (x \big|_{1}^{5}) = \ln 2 \cdot (5 - 1) = 4 \ln 2.\]

Step 6: Present the final result

The area between the curves \( y = \ln x \) and \( y = \ln(2x) \) from \( x = 1 \) to \( x = 5 \) is \( 4 \ln 2 \).

Key Concepts

Area Between Curves

Finding the area between two curves is a common problem in calculus, and it helps determine the space enclosed by these curves on a given interval. This can be visually understood as the 'shaded' region between the lines on a graph.
To find this area, we first identify which curve is above the other within the interval. This helps set the bounds correctly from where to integrate. In this specific problem, the function

• \( y = \ln(2x) \) is always above \( y = \ln x \) on the interval from \( x = 1 \) to \( x = 5 \) because \( \ln(2x) \) simplifies to \( \ln x + \ln 2 \).

Once the functions and the bounds are identified, the area between the curves is calculated using the definite integral of the difference between these two functions.We derive the integrand by finding the difference:

• \( \ln(2x) - \ln x = \ln 2 \).

This consistent difference shows that the distance between these curves is constant over the interval, which simplifies the problem greatly.

Integration

Integration is a fundamental concept in calculus which allows us to calculate things like areas under curves, among other applications. It's essentially a way of summing up infinitesimally small quantities to get a total.
In the context of this problem, integration helps us find the total area between the two curves over a specified interval.For our problem, the integral set up is:

• \[\text{Area} = \int_{1}^{5} [\ln(2x) - \ln x] \, dx\].

However, after simplifying, the expression inside the integral simplifies to:

• \( \ln 2 \), meaning we are essentially integrating a constant over the interval \( x = 1 \) to \( x = 5 \).

When integrating a constant, you multiply the constant by the difference in bounds, which makes it much easier to solve compared to more complex integrals. This is because integrating a constant

• \( C \) over the interval \( a \) to \( b \) simply results in \( C(b-a) \).

In this case, the result is:

• \( 4 \ln 2 \).

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a logarithm with the base \( e \), where \( e \approx 2.71828 \). It appears frequently in calculus and complex mathematical calculations.
In our exercise, both functions include the natural logarithm. The first is \( y = \ln x \), and the second is \( y = \ln(2x) \). A useful property of logarithms simplifies our problem substantially: the addition of logarithms, such as \( \ln(2x) \), can be rewritten using logarithmic rules as:

• \( \ln(2x) = \ln 2 + \ln x \).

By understanding this property, we directly see that \( \ln(2x) \) is consistently above \( \ln x \) by exactly \( \ln 2 \) across our interval from \( x = 1 \) to \( x = 5 \). This transforms what could have been a complicated comparison into a simple and direct subtraction of the two functions, leading to a straightforward integration.Understanding these properties not only aids in simplifying the current calculation but also provides tools for tackling more complex problems in calculus.