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Magazine Chapter 7: Problem 65
Find the area of the region bounded by the curve \(y=\ln x,\) the \(x\) -axis, and the line \(x=e\).
Short Answer
Expert verified
The area is 1 square unit.
Step by step solution
01
Understand the Problem
We need to find the area between the curve given by the function \(y = \ln x\), the \(x\)-axis, and the vertical line \(x = e\). Since \(x\) is positive on these intervals, we observe the curve from where \(x\) starts at the point it crosses the \(y\)-axis, which is \(x=1\), up to \(x=e\).
02
Determine the Limits of Integration
The region is bounded from \(x=1\) to \(x=e\) because the natural logarithm \(y=\ln x\) crosses the \(y\)-axis at \(x=1\), and the problem specifies \(x=e\) as the other boundary.
03
Set Up the Integral
The area under the curve \(y = \ln x\) from \(x=1\) to \(x=e\) can be calculated using the definite integral: \[\int_{1}^{e} \ln x \, dx.\]
04
Use Integration by Parts
To integrate \(\int \ln x \, dx\), we use the integration by parts formula: \[\int u \, dv = uv - \int v \, du.\]Let \(u = \ln x\) and \(dv = dx\). Then, \(du = \frac{1}{x} dx\) and \(v = x\).
05
Evaluate Integration by Parts
Substituting into the integration by parts 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 becomes the integral \[[ x \ln x - x ]\_{1}^{e}.\]
06
Calculate the Definite Integral
Substituting the limits \(x=1\) and \(x=e\) into the expression \(x \ln x - x\), we get:\[\begin{align*}[(e \ln e - e) - (1 \ln 1 - 1)] &= (e \cdot 1 - e) - (0 - 1) \&= (e - e + 1) = 1.\end{align*}\]
07
Final Result
The area of the region bounded by the curve \(y=\ln x\), the \(x\)-axis, and the line \(x=e\) is 1 square unit.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by parts
Integration by parts is a method that is particularly useful for integrating products of functions. It's based on the product rule for differentiation, but applied in reverse. When you encounter an integral like \( \int u \, dv \), you can use the formula:\[\int u \, dv = uv - \int v \, du\]Here's how it works for \( \int \ln x \, dx \):
- Identify parts of the integrand as \( u \) and \( dv \). For \( \ln x \), let \( u = \ln x \) and \( dv = dx \).
- Differentiate \( u \) to get \( du = \frac{1}{x} dx \), and integrate \( dv \) to get \( v = x \).
- Substitute these into the integration by parts formula to simplify the integral.
- The resulting integral \( \int x \cdot \frac{1}{x} \, dx = \int 1 \, dx = x \) is straightforward, simplifying the problem considerably.
Definite integral
The concept of a definite integral is central to calculating the area under a curve. A definite integral has limits of integration, which specify the interval over which we calculate the area. In this problem, we are looking at the curve \( y = \ln x \) from \( x=1 \) to \( x=e \). The corresponding definite integral is:\[\int_{1}^{e} \ln x \, dx\]
- The lower limit of 1 indicates where the curve meets the vertical axis at the point \( x = 1 \), as the natural logarithm of 1 is zero.
- The upper limit is \( x=e \), given as part of the problem to find the bounded region.
Natural logarithm
The natural logarithm, denoted as \( \ln x \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It's a fundamental function in calculus due to its unique properties. When plotted, the curve \( y = \ln x \) crosses the \( y \)-axis at \( x = 1 \). This point is relevant because:
- The function \( \ln x \) is defined for all positive \( x \), with \( \ln 1 = 0 \).
- This characteristic makes \( x=1 \) a pivotal starting point when calculating areas involving \( \ln x \).
