Question

. Find the area of the region bounded by the curve \(y=\ln x\), the \(x\) -axis, and the line \(x=e .\)

Short answer

The area is 1.

Step-by-step solution

Understand the Region to be Integrated

The problem asks us to find the area bounded by the curve \(y = \ln x\) above the \(x\)-axis, and between the lines \(x = 1\) and \(x = e\). Here, the function \(y = \ln x\) represents a natural logarithm curve starting from the point \((1, 0)\)

Set Up the Definite Integral

To find the area under the curve \(y = \ln x\), we need to set up the definite integral from \(x = 1\) to \(x = e\). The integral to solve is: \[ \int_1^e \ln x \, dx. \]

Integrate the Function

Use integration by parts to solve the integral \(\int \ln x \, dx\). Let \(u = \ln x\) and \(dv = dx\). Then, \(du = \frac{1}{x}dx\) and \(v = x\). By the integration by parts formula, \(\int u \, dv = uv - \int v \, du,\) we find that:\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C. \]

Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative obtained from Step 3:\[ \int_1^e \ln x \, dx = \left[x \ln x - x \right]_1^e. \]Calculate the value:- When \(x = e\): \(e \ln e - e = e(1) - e = e - e = 0.\)- When \(x = 1\): \(1 \cdot \ln 1 - 1 = 0 - 1 = -1.\)Thus, the area is \(0 - (-1) = 1.\)

Conclusion

The area of the region bounded by \(y = \ln x\), the \(x\)-axis, and the line \(x = e\) is 1. This makes sense as the graph of \(y = \ln x\) between \(x=1\) and \(x=e\) is concave down, starting at zero when \(x=1\) and asymptotically reaching zero there.

Key concepts

Definite Integral

The definite integral is a fundamental concept in calculus used to calculate the area under a curve. This is exactly what we're doing when we determine the area bounded by a curve and specified lines. In the specific problem we are discussing, we aim to find the area under the curve \( y = \ln x \) from \( x = 1 \) to \( x = e \). This involves setting up an integral with precise limits, which are the boundaries of the region.
Remember, the definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_a^b f(x) \ dx \). It represents the net area between the function and the \(x\)-axis, considering areas above the axis as positive and below as negative.
Here's a brief overview of the process:

• Select the function whose area needs to be computed.
• Set the upper and lower limits, which here are \( x = e \) and \( x = 1 \), respectively.
• Compute the integral over these specified limits to find the enclosed area.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a logarithmic function with the base \( e \), the Euler's number which is approximately 2.71828.
It has several interesting properties that come into play during integration, specifically when used with functions like the one in this problem \( y = \ln x \).
The natural logarithm increases as \( x \) increases, but at a decreasing rate, reflecting its distinctive curve shape. At \( x = 1 \), \( \ln(1) = 0 \); this point is key as it represents where the curve intersects the \( x\)-axis in our bounded region.
Some important properties of the natural logarithm are:

• \( \ln(e) = 1 \)
• \( \ln(ab) = \ln a + \ln b \)
• \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• \( \ln(1) = 0 \)

Understanding these properties helps simplify the integration, as correctly identifying the behavior of \( \ln x \) allows us to set accurate limits and evaluate resultant expressions.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is particularly handy when dealing with functions like \( \ln x \) within an integral, as it's difficult to integrate directly.
The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \]
This formula arises from the product rule of differentiation and allows us to break an integral involving a product of functions into simpler parts. In our exercise, we use integration by parts to evaluate \( \int \ln x \, dx \).Here's the breakdown of how this technique works:

• Choose parts: \( u = \ln x \) and \( dv = dx \), from which \( du = \frac{1}{x} \, dx \) and \( v = x \).
• Apply the formula: \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C \).

This result is then used to evaluate our definite integral to find the area bounded by the curve. By substituting the bounds of integration, this expression simplifies the process of evaluating the definite integral and finding the precise area.

Bounded Region

The concept of a bounded region is essential when discussing definite integrals and finding areas. A bounded region refers to the space surrounded by curves, lines, or axes, creating a closed area whose measure we can calculate using integration.
In our problem, we deal with a region bounded by:

• The curve \( y = \ln x \)
• The \( x\)-axis
• The vertical lines \(x = 1\) and \( x = e \)

This setup creates a clear boundary that encloses an area, manageable within the limits of integration corresponding to the coordinates where the curve meets these lines.
For a successful computation, identifying and clearly understanding these boundaries is crucial. It ensures that when the integral is evaluated, it only considers the desired area. The process of setting limits in the definite integral precisely reflects these boundaries, allowing for accurate measurement of the area.