Problem 71 - Thomas Calculus

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 \(4 \ln 2\) or approximately 2.772.

Step 1: Understand the Problem

We need to find the area between two curves, specifically \(y = \ln x\) and \(y = \ln (2x)\), over the interval \(x = 1\) to \(x = 5\). The area can be found by calculating the definite integral of the difference of the two functions over this interval.

Step 2: Set Up the Integral

The function \(y = \ln 2x\) can be rewritten as \(y = \ln 2 + \ln x\) using logarithmic properties. Hence, the difference between the functions for any \(x\) is \(\ln 2x - \ln x = \ln 2 + \ln x - \ln x = \ln 2\). Therefore, the integral simplifies to \(\int_{1}^{5} \ln 2\, dx\).

Step 3: Compute the Integral

The integral \(\int_{1}^{5} \ln 2\, dx\) can be simplified since \(\ln 2\) is a constant factor. The integral becomes \(\ln 2 \times \int_{1}^{5} 1\, dx\). The integral of 1 with respect to \(x\) over the interval \([1,5]\) is \(x\Big|_1^5\), which equals \(5 - 1 = 4\). Thus, our expression simplifies to \(4 \times \ln 2\).

Step 4: Calculate the Area

Now, multiplying the result by the constant \(\ln 2\), we get the area as \(4 \times \ln 2\). Converting to a decimal approximation, this area is approximately \(4 \times 0.693 = 2.772\). Thus, the exact area between the curves is \(4 \ln 2\), which is approximately 2.772.

Key Concepts

Definite Integral

A definite integral is a crucial concept in calculus used to find the accumulation of quantities, such as the area under a curve.
When dealing with functions that intersect or form enclosed areas, we can use definite integrals to calculate the area between them. In the problem we solved, we dealt with two logarithmic functions over a set interval from 1 to 5.
The definite integral \[ \int_{1}^{5} \ln 2 \, dx \] is computed over the interval \([1, 5]\),
where \( \ln 2 \) is a constant. Because the difference function between the two curves was constant, the integral simplified greatly.

• It's important to remember that when integrating a constant, we treat it like any other constant multiplication.
• The integral helps us find the net area between the two curves.

It effectively subtracts the value of one logarithmic function from the other over the given interval.

Logarithmic Functions

Logarithmic functions are based on exponential relationships and have unique properties that make them interesting to work with in calculus.
In our problem, we encountered two such functions: \( y = \ln x \) and \( y = \ln (2x) \).

• A key property of logarithms is that they can be added due to the multiplication inside the log: \( \ln(2x) = \ln 2 + \ln x \).
• This property allowed us to express \( \ln (2x) \) in a simpler form, which was instrumental in finding the difference between the two curves as a simple constant, \( \ln 2 \).

Understanding this transformation simplifies the calculation and helps see that the curves differ by a constant amount.

Integral Calculus

Integral calculus involves the study of integrals and their properties. It often deals with finding areas, volumes, central points, and many useful quantities.
The most straightforward application is calculating areas under curves, as in our problem. Calculating the area between two curves is a specific case where integral calculus shines.

• Integration as the reverse of differentiation: While derivatives give the rate of change, integrals can return us to the original function or give the total accumulation.
• By employing integral calculus, we take advantage of its ability to handle continuous change and dissect the differences between two continuous functions over a defined interval.

In the original problem, \( \int_{1}^{5} \ln 2 \, dx \) represents a straightforward application where the integration of a constant factor is performed, leading directly to the area calculation.