Textbook Question
Finding volume: Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y = e^(-x), and the line x = 1.
a. About the y-axis.
Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.2.63a
Consider the region bounded by the graphs of
y = ln(x), y = 0, and x = e.
a. Find the area of the region.
Verified step by step guidance1
Identify the region bounded by the curves: the graph of \(y = \ln(x)\), the line \(y = 0\), and the vertical line \(x = e\). Note that \(y = 0\) is the x-axis.
Determine the interval for \(x\) over which the region lies. Since \(y = \ln(x)\) intersects \(y = 0\) when \(\ln(x) = 0\), solve for \(x\): \(\ln(x) = 0 \implies x = 1\). So the region is bounded between \(x = 1\) and \(x = e\).
Set up the integral for the area between the curves. The area \(A\) is given by the integral of the top function minus the bottom function over the interval \([1, e]\). Here, the top function is \(y = \ln(x)\) and the bottom function is \(y = 0\), so:
\[ A = \int_{1}^{e} (\ln(x) - 0) \, dx = \int_{1}^{e} \ln(x) \, dx \]
To evaluate the integral \(\int \ln(x) \, dx\), use integration by parts. Let \(u = \ln(x)\) and \(dv = dx\). Then, \(du = \frac{1}{x} dx\) and \(v = x\). Apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Substitute back to find the definite integral from 1 to \(e\) and express the area in terms of these evaluated expressions.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals and Area Under a Curve
The definite integral of a function over an interval represents the net area between the curve and the x-axis. When the function is positive, the integral gives the exact area under the curve. Calculating the area bounded by curves often involves setting up and evaluating definite integrals.
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Definition of the Definite Integral
Natural Logarithm Function Properties
The natural logarithm function, y = ln(x), is defined for x > 0 and is the inverse of the exponential function e^x. It is continuous and increasing, with ln(1) = 0. Understanding its behavior helps in setting integration limits and interpreting the region bounded by y = ln(x), y = 0, and vertical lines.
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06:21Properties of Functions
Setting Integration Limits from Boundary Curves
To find the area of a region bounded by curves, it is essential to identify the correct limits of integration. These limits correspond to the intersection points of the bounding curves. In this problem, the vertical boundary x = e and the x-values where y = ln(x) meets y = 0 determine the integration interval.
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Related Practice
Textbook Question
Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 22,000 miles and σ = 4,000 miles.
a. In a batch of 4000 tires, how many can be expected to last for at least 18,000 miles?
1
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Textbook Question
81. Find the values of p for which each integral converges.
a. ∫ from 1 to 2 of [dx / (x (ln x)^p)]
Textbook Question
Using different substitutions
Show that the integral
∫((x² - 1)(x + 1))^(-2/3) dx
can be evaluated with any of the following substitutions.
a. u = 1/(x + 1)
What is the value of the integral?
Textbook Question
Finding area
Find the area of the region enclosed by the curve y = x cos(x) and the x-axis (see the accompanying figure) for:
a. π/2 ≤ x ≤ 3π/2.
1
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Textbook Question
In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (a) the Trapezoidal Rule (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from -1 to 1 of (x² + 1) dx