Problem 6 - Thomas Calculus

Question

In Exercises 1–8, sketch the region bounded by the given lines and curves. Then express the region’s area as an iterated double integral and evaluate the integral. The curves \(y=\ln x\) and \(y=2 \ln x\) and the line \(x=e,\) in the first quadrant

Step-by-Step Solution

Answer

The area of the region is 1 square unit.

Step 1: Understand the curves and region

We need to sketch the curves given by the equations. The curve \(y = \ln x\) represents a logarithmic function that passes through the point \((1, 0)\) and curves upwards. The curve \(y = 2 \ln x\) also passes through \((1, 0)\) but grows faster than \(y = \ln x\). The line \(x = e\) is a vertical line passing through the point where \(x = e\), approximately \(2.718\). We are considering all these in the first quadrant where all values are positive.

Step 2: Determine the region of interest

Since we are interested in the first quadrant, the region of interest is bounded above by the curve \(y = 2 \ln x\), below by \(y = \ln x\), and to the right by the line \(x = e\). We consider \(x = 1\) as the left boundary for the area since both curves \(y = \ln x\) and \(y = 2\ln x\) meet at \(x=1\). Thus, our region is bounded by \(1 \leq x \leq e\).

Step 3: Set up the iterated double integral

The area between two curves can be determined by integrating the difference of the functions over the bounds. The bounds for \(x\) are from 1 to \(e\). The function \(y = 2\ln x\) is greater than \(y = \ln x\) for all \(x\) values from 1 to \(e\). Thus, the iterated integral for the area is:\[A = \int_{1}^{e} \int_{\ln x}^{2 \ln x} dy \, dx\]

Step 4: Evaluate the inner integral

Evaluate the inner integral with respect to \(y\):\[\int_{\ln x}^{2 \ln x} dy = \left[ y \right]_{\ln x}^{2 \ln x} = (2 \ln x) - (\ln x) = \ln x\]

Step 5: Evaluate the outer integral

Now, evaluate the outer integral with respect to \(x\):\[A = \int_{1}^{e} \ln x \, dx\]To solve this, use integration by parts, where \(u = \ln x\) and \(dv = dx\).- \(du = \frac{1}{x} dx\)- \(v = x\)The integration by parts formula is \(\int u \, dv = uv - \int v \, du\):\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx\]\[= x \ln x - \int 1 \, dx = x \ln x - x + C\]Evaluate from 1 to \(e\):\[[x \ln x - x]_{1}^{e} = \left[e \ln e - e\right] - \left[1 \ln 1 - 1\right] = [e - e] - [0 - 1] = 0 + 1 = 1\]

Step 6: Conclusion

The area bounded by the curves \(y = \ln x\) and \(y = 2 \ln x\) between \(x = 1\) and \(x = e\) is 1 square unit.

Key Concepts

Logarithmic Functions

Logarithmic functions, represented mathematically as functions of the form \( y = \ln x \), are the inverses of exponential functions. The natural logarithm, \( \ln x \), uses the constant \( e \) (approximately 2.718) as its base. This function is defined only for positive values of \( x \), reflecting the fact that you cannot take the logarithm of zero or a negative number in the real number system.
For the curve \( y = \ln x \), it smoothly passes through the point \((1, 0)\) since \( \ln(1) = 0 \). Both \( y = \ln x \) and its scaled variant \( y = 2\ln x \) are types of logarithmic functions. The latter grows faster because multiplying the logarithm by 2 stretches the curve vertically by a factor of 2.
These properties are critical for understanding their behavior in relation to each other and to boundaries like \( x = e \). When compared, \( y = 2\ln x \) is always above \( y = \ln x \) when \( x > 1 \), which is crucial for identifying the area between these curves.

Area Between Curves

The concept of finding the area between curves is a fundamental application of integration. In the case of the given exercise, we are examining the region between the curves \( y = \ln x \) and \( y = 2 \ln x \) over an interval. The curves are further bounded by the lines \( x = 1 \) and \( x = e \), simplifying our region of interest.
To calculate the area between these two curves:

• Determine the difference between the two functions: \( (2 \ln x) - (\ln x) = \ln x \)
• Integrate this difference over the interval from \( x = 1 \) to \( x = e \)

By doing so, we effectively compute the vertical "slice" of area at every point in the interval and accumulate those to find the total area. This process leads us to an expression of the area as a double integral, which is crucial in evaluating the space between the overlapping functions.

Iterated Integrals

Iterated integrals are a technique used to evaluate multiple integrals, particularly in the context of calculating areas and volumes. An iterated integral arises when you first solve an inner integral, and the result serves as the integrand for the outer integral. This technique simplifies the computation process by breaking down a complex multi-dimensional integral into simpler, sequential one-dimensional integrals.
In our exercise, we use an iterated double integral to find the area between two curves. We first integrate with respect to \( y \) over the limits \( y = \ln x \) to \( y = 2 \ln x \). The result, \( \ln x \), is then integrated with respect to \( x \) from 1 to \( e \). This two-step process is an example of iterated integration:

• First Integral: \( \int_{\ln x}^{2 \ln x} dy \)
• Second Integral: \( \int_{1}^{e} (\text{result of first integral}) \, dx \)

Using iterated integrals not only streamlines the calculation but also provides a systematic approach to tackling complex integration problems. The final result of this process is a quantitative measurement of the area, which in this case turns out to be 1 square unit.