Problem 73 - 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 \(4 \ln 2\).

Step 1: Identify the Functions and Bounds

We need to find the area between the curves given by the functions \(y = \ln x\) and \(y = \ln 2x\), within the interval \([1, 5]\).

Step 2: Express the Area as an Integral

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

Step 3: Simplify the Integral

Simplify \(\ln (2x) - \ln x\) using the properties of logarithms: \[\ln (2x) - \ln x = \ln \left(\frac{2x}{x}\right) = \ln 2\]So the integral becomes:\[\text{Area} = \int_{1}^{5} \ln 2 \, dx\].

Step 4: Compute the Integral

Since \(\ln 2\) is a constant, the integral of \(\ln 2\) from 1 to 5 is:\[\text{Area} = \ln 2 \int_{1}^{5} dx = \ln 2 [x]_{1}^{5}\].

Step 5: Evaluate the Definite Integral

Evaluate the expression:\[\text{Area} = \ln 2 \times (5 - 1) = \ln 2 \times 4 = 4 \ln 2\].

Key Concepts

Definite Integral

The concept of a definite integral allows us to find the exact area under a curve between two points on the x-axis. In this problem, the definite integral is used to find the area between two curves, specifically from the functions \(y = \ln x\) and \(y = \ln 2x\) between \(x = 1\) and \(x = 5\). To set up the definite integral, we subtract the lower curve from the upper curve and integrate over the specified interval. This gives us the expression \(\int_{1}^{5} (\ln (2x) - \ln x) \, dx\). Breaking down a problem in this way simplifies calculating areas where functions intersect or integrate over different sections.

Logarithmic Functions

Logarithmic functions, such as \(\ln x\), play a crucial role in calculus and represent a natural way to express exponential relationships. The log function \(\ln x\) is the natural logarithm, which uses the base \(e\), an irrational number approximately equal to 2.718. When dealing with areas between curves, as in this problem, logarithmic functions can simplify to constants, as we will see through the use of their properties. By understanding these functions well, one can easily manipulate and simplify integrals involving them, ultimately making the calculation process smoother and more intuitive.

Properties of Logarithms

The properties of logarithms are powerful tools for simplifying expressions like the one you see in this problem. Here, we use the property \(\ln ab = \ln a + \ln b\), and more importantly, \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\). In this exercise, \(\ln (2x) - \ln x\) is simplified to \(\ln \left(\frac{2x}{x}\right) = \ln 2\). This drastic simplification from a variable expression to a constant, \(\ln 2\), enables the straightforward integration over the defined interval. Knowing how to apply these properties, one can simplify what appears to be complex problems into easier computations, enhancing both understanding and problem-solving efficiency.