Problem 51 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Use a double integral to find the area of \(R\).\(R\) is the region bounded by \(y=\ln x, y=0\), and \(x=e\).

Step-by-Step Solution

Answer

The area of region R is 1.

Step 1 - Understand the Region

Identify the boundaries of the region R. The region is bounded by the curve \(y = \ln x\), the line \(y = 0\) (the x-axis), and the vertical line \(x = e\).

Step 2 - Set Up the Double Integral

To find the area, set up a double integral. For the given region, the bounds for x are from \(1\) to \(e\) because \(\ln(1) = 0\) and \(\ln(e) = 1\). The y-values will range from \(0\) to \(\ln x\). The double integral is \[A = \int_{1}^{e} \int_{0}^{\ln x} \, dy \, dx.\]

Step 3 - Integrate with Respect to y

Integrate the inner integral with respect to y. \[\int_{0}^{\ln x} \, dy = y \Big|_{0}^{\ln x} = \ln x.\]

Step 4 - Integrate with Respect to x

Substitute the result of the inner integral into the outer integral and integrate with respect to x. \[A = \int_{1}^{e} \ln x \, dx\] To solve this, use integration by parts where we let \(u = \ln x\) and \(dv = dx\), giving \(du = \frac{1}{x}dx\) and \(v = x\).

Step 5 - Apply Integration by Parts

Applying integration by parts, we get: \[ A = x \ln x \Big|_{1}^{e} - \int_{1}^{e} x \cdot \frac{1}{x} \, dx = e \ln e - 1 \ln 1 - \int_{1}^{e} 1 \, dx.\] Since \(\ln e = 1\) and \(\ln 1 = 0\), this simplifies to \[A = e - \int_{1}^{e} 1 \, dx = e - (e - 1) = 1.\]

Step 6 - Final Area Calculation

So, our final area is \[A = 1.\]

Key Concepts

Area under the curve

The area under a curve is a fundamental concept in calculus used to determine the size of a region bounded by the curve and the x-axis. To find this area, we often use integration. In this exercise, the region R is bounded by the natural logarithm function, the x-axis, and a vertical line at x = e. This requires the use of a double integral, which helps in summing up small areas to find the total area of the region.

We first visualize the bounds:

• The curve is \(y = \ln x\).
• The x-axis is \(y = 0\).
• The vertical line at x = e marks the right boundary.

To sum up all the individual small areas within this region, we set up a double integral which will account for the area in both x and y directions.

In our example, we set the x-bounds from 1 to e because at x = 1, \(\text{ln} 1 = 0\) and at x = e, \(\text{ln} e = 1\). Subsequently, the y-values range from the x-axis (y=0) to the curve \(y = \ln x\). This leads to the integral: \[ A = \int_{1}^{e} \int_{0}^{\text{ln} x} dy \, dx. \]By calculating this integral step by step, we find the area under the curve in region R.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is crucial when facing certain integrals that are not straightforward.

The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du. \]
In our exercise, after integrating the inner integral, we need to compute \( \int_{1}^{e} \text{ln} x dx \). This is where integration by parts becomes useful. We choose:

• \