Problem 71 - Thomas Calculus in SI Units

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

Area = \(4 \ln 2\), approximately 2.772 square units.

Step 1: Determine Points of Intersection

First, identify the range of integration by solving for points of intersection between the two curves. Set the equations equal to each other: \( \ln x = \ln 2x \). Simplify this equation:\[ \ln x = \ln 2 + \ln x \]This simplifies to:\[ 0 = \ln 2 \]This is not possible, indicating that the graphs do not intersect within the given bounds, thus we proceed to integrate over the given interval.

Step 2: Determine the Functions

Determine which curve is on top between \( x = 1 \) and \( x = 5 \). Since \( \ln 2x = \ln 2 + \ln x \), which is \( \ln x + \ln 2 \), it is clear that for any \( x > 0 \), \( \ln 2x \) is greater than \( \ln x \). Hence, \( y = \ln 2x \) is the upper function.

Step 3: Set up the Integral

The area between the two curves is given by integrating the difference of the two functions. Set up the definite integral:\[ \int_{1}^{5} ((\ln 2x) - (\ln x)) \, dx \]

Step 4: Simplify the Integrand

Simplify the expression inside the integral:\[ (\ln 2x) - (\ln x) = \ln 2 + \ln x - \ln x = \ln 2 \]Thus, the integral becomes:\[ \int_{1}^{5} \ln 2 \, dx \]

Step 5: Calculate the Integral

The integral simplifies to:\[ \ln 2 \int_{1}^{5} 1 \, dx = \ln 2 [x]_{1}^{5} \]Evaluate the definite integral:\[ \ln 2 (5 - 1) = \ln 2 \times 4 = 4 \ln 2 \]

Step 6: Concluding the Calculation

Calculate the result in numerical terms, if necessary, using \( \ln 2 \approx 0.693 \). Thus, the area is approximately:\[ 4 \times 0.693 = 2.772 \] square units.

Key Concepts

Definite Integral

In calculus, a definite integral helps us find the area between two curves over a specific interval on the x-axis. This concept is essential for calculating the exact area rather than just estimating. When working with functions, the definite integral is set up as an integral from one boundary to another. For our exercise, this means integrating from \( x=1 \) to \( x=5 \). The basic idea is to calculate the exact accumulated area by comparing the top curve against the bottom curve over the given interval. This involves finding the difference between two functions, \( y = \ln 2x \) and \( y = \ln x \). By subtracting these two within a definite integral, we can compute our desired area. This concept can be depicted mathematically as:

• \(\int_{a}^{b} (f(x) - g(x)) \, dx\)

Here, \( f(x) \) is the upper function and \( g(x) \) is the lower one. In our case, the integrand simplifies down to \( \ln 2 \), allowing us to focus on the limits of integration to find the area.

Natural Logarithm

The natural logarithm, represented as \( \ln \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. It is used frequently in calculus due to its simple derivative and integral properties. In this problem, we have two functions involving natural logarithms: \( y = \ln x \) and \( y = \ln 2x \). Notice how neat it is when applying logarithmic rules like \( \ln(ab) = \ln a + \ln b \). So, \( \ln 2x \) becomes \( \ln 2 + \ln x \).This reformulation is crucial since it shows why \( \ln 2x \) is greater than \( \ln x \) by a constant value of \( \ln 2 \). When moving through problems like this one, being comfortable with logarithm identities can help simplify complex expressions and ultimately aid in finding intersections or areas.

Points of Intersection

Determining the points of intersection between curves helps you establish boundaries where the functions meet. In our problem, you set the two functions equal to each other to find such points: \[ \ln x = \ln 2x \]This simplifies to \[ \ln x = \ln 2 + \ln x \]Then, you get:\[0 = \ln 2\]Here, since \( \ln 2 \) is a constant, it indicates that the equations do not intersect in the given interval from \( x = 1 \) to \( x = 5 \). Knowing whether intersections exist is important, as they often mark the boundaries across which you integrate. In scenarios where curves do intersect, these would become limits for setting up the integral. Careful examination and simplification help reveal whether intersections exist, keeping calculations on track and ensuring an accurate evaluation of the area.