Textbook Question
88. The region in Exercise 87 is revolved about the x-axis to generate a solid.
a. Find the volume of the solid.
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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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.2.57b
Finding area
Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:
b. π ≤ x ≤ 2π.
Verified step by step guidance1
Identify the region whose area is to be found: the curve is given by \(y = x \sin x\) and the interval is \(\pi \leq x \leq 2\pi\). From the graph, note that the curve lies below the x-axis in this interval, so the function values are negative.
Set up the integral for the area between the curve and the x-axis. Since the curve is below the x-axis, the area is given by the integral of the negative of the function: \(\text{Area} = \int_{\pi}^{2\pi} -x \sin x \, dx\).
Use integration by parts to evaluate the integral \(\int x \sin x \, dx\). Let \(u = x\) and \(dv = \sin x \, dx\), then \(du = dx\) and \(v = -\cos x\). Applying integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
Substitute back into the definite integral and evaluate the resulting expression at the limits \(x = \pi\) and \(x = 2\pi\).
Take the absolute value of the result if necessary to ensure the area is positive, as area cannot be negative.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integral for Area Calculation
The definite integral of a function over an interval gives the net area between the curve and the x-axis. When the function dips below the x-axis, the integral yields a negative value, so the absolute value or splitting the integral at zeros is necessary to find the total enclosed area.
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Definition of the Definite Integral
Behavior of the Function y = x sin(x)
The function y = x sin(x) oscillates with increasing amplitude as x increases. It crosses the x-axis at multiples of π, creating regions above and below the axis. Understanding where the function is positive or negative helps determine how to set up the integral for area.
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03:39Integrals of Natural Exponential Functions (e^x)
Splitting the Integral at Zeros of the Function
To find the area between the curve and the x-axis over an interval where the function changes sign, split the integral at points where y = 0. Calculate the integral over each subinterval and take the absolute value of each to sum the total area.
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Related Practice
Textbook Question
Evaluate ∫ x³ √(1 - x²) dx using:
a. Integration by parts.
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 -2 to 0 of (x² - 1) dx
Textbook Question
89. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / x², y = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (i) about the x-axis.
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Textbook Question
Finding volume: Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes and the curve y = cos(x), 0 ≤ x ≤ π/2, about
b. The line x = π/2.
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 (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from 0 to 3 of 1/√(x + 1) dx