Problem 65 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The area is 1.

Step 1: Understand the Problem

We need to find the area under the curve \(y = \ln x\) from \(x = 1\) to \(x = e\). This means we are integrating \(\ln x\) with respect to \(x\) over this interval.

Step 2: Set Up the Integral

The area under the curve between \(x = 1\) and \(x = e\) is given by the integral \[ \int_{1}^{e} \ln x \, dx. \]

Step 3: Integrate the Function

To integrate \(\ln x\), we use integration by parts. Let \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} dx\) and \(v = x\). The integration by parts formula is \(\int u \, dv = uv - \int v \, du.\)Applying the formula:\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx. \]

Step 4: Solve the Integral

Now solve \(x \ln x - \int 1 \, dx\):\[ = x \ln x - x + C. \]We obtain the antiderivative of \(\ln x\) as \(x \ln x - x.\)

Step 5: Evaluate the Definite Integral

Substitute the limits into the antiderivative:\[ \left[ x \ln x - x \right]_1^e = \left( e \ln e - e \right) - \left(1 \ln 1 - 1\right). \]

Step 6: Simplify the Expression

Evaluate the expression:\[ e \ln e - e = e(1) - e = 0, \]and\[ 1 \ln 1 - 1 = 0 - 1 = -1. \]

Step 7: Compute the Final Result

The area is \[ 0 - (-1) = 1. \] This is the area of the region under the curve from \(x = 1\) to \(x = e\).

Key Concepts

Area Under the Curve

Finding the area under a curve is an essential concept in calculus. This involves using the integral calculus, which allows us to measure the total accumulation of quantities. In the problem, we need to find the area beneath the curve of the natural logarithm function, \(y = \ln x\), from \(x = 1\) to \(x = e\).
This area is essentially the accumulation of all the tiny areas under the curve between these two points on the x-axis.

• The boundaries are set by the limits of integration, here being \(x=1\) and \(x=e\).
• Instead of attempting to measure this area by hand-drawing, integrals allow for precise calculation.
• The result represents the total area, which, in this case, helped determine the space under \(y = \ln x\).

Understanding this process helps visualize how much 'space' the curve covers against the x-axis between those given points.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It's especially helpful when you encounter functions that are under a logarithmic, exponential, or polynomial form. In our exercise, integrating \(\ln x\) with respect to \(x\) requires this method because direct integration isn't straightforward.
The integration by parts formula is: \(\int u\, dv = uv - \int v\, du\).

• To use this, we select \(u\) and \(dv\). Here, \(u = \ln x\) and \(dv = dx\).
• This choice leads to \(du = \frac{1}{x} dx\) and \(v = x\).

Once these substitutions are set, plugging them into the formula helps break down the integral into smaller and more manageable pieces. Breaking down the problem like this makes it easier to arrive at the solution.

Definite Integrals

Definite integrals give us the ability to calculate the exact area under a curve between two points on the x-axis. This is different from indefinite integrals, which yield a family of functions. In our exercise, the integral: \[ \int_{1}^{e} \ln x \, dx \]

• The definite integral takes bounds, here from \(1\) to \(e\).
• After applying the integration by parts method, the expression becomes: \[ x \ln x - \int 1 \, dx.\]

After obtaining the antiderivative, \(x \ln x - x\), we substitute the limits of integration, plugging \(e\) for \(x\) and subsequently \(1\). Subtracting these results yields the exact area beneath the curve over the specified interval.

Natural Logarithm Function

The natural logarithm function, denoted as \(\ln x\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. In calculus, it's a widely used function due to its smooth and continuous characteristics.
In this problem, \(y = \ln x\), we're dealing with the curve that shows a steady increase and represents the natural growth processes.

• The function is defined only for positive \(x\), making \(x = 0\) a vertical asymptote.
• The derivative of \(\ln x\) is \(\frac{1}{x}\), which signifies the rate of change of this curve.

Understanding the behavior of the natural logarithm is key to performing calculus operations with it. It is especially useful in integration, particularly when you employ techniques such as integration by parts, as displayed in our exercise.