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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.2.58b
Chapter 8, Problem 8.2.58b

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:
b. 3π/2 ≤ x ≤ 5π/2.
Graph of y = x cos(x) from 0 to 7π/2 showing oscillations crossing the x-axis between 3π/2 and 5π/2.

Verified step by step guidance
1
Identify the region bounded by the curve \(y = x \cos(x)\) and the x-axis for the interval \(\frac{3\pi}{2} \leq x \leq \frac{5\pi}{2}\). Notice from the graph that the curve crosses the x-axis at \(x = \frac{3\pi}{2}\) and \(x = \frac{5\pi}{2}\), and the function is negative between these points.
Since the function is below the x-axis in this interval, the area between the curve and the x-axis is given by the integral of the absolute value of the function. This means we need to integrate \(-y = -x \cos(x)\) over \(\left[ \frac{3\pi}{2}, \frac{5\pi}{2} \right]\) to get a positive area.
Set up the integral for the area as \(\displaystyle A = \int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}} -x \cos(x) \, dx\).
Use integration by parts to evaluate the integral. Let \(u = x\) and \(dv = \cos(x) dx\). Then, compute \(du = dx\) and \(v = \sin(x)\). Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
After finding the antiderivative, evaluate it at the bounds \(x = \frac{5\pi}{2}\) and \(x = \frac{3\pi}{2}\), then subtract to find the definite integral value, which represents the area of the region.

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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 crosses the x-axis, the integral sums positive and negative areas, so to find the total enclosed area, you must consider the absolute value or split the integral at zeros.
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Definition of the Definite Integral

Finding Zeros of the Function

To determine the area enclosed by the curve and the x-axis, it is essential to find where the function crosses the x-axis (i.e., where y = 0). For y = x cos(x), zeros occur when cos(x) = 0 or x = 0, which helps in splitting the integral into intervals where the function is positive or negative.
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Guided course
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Zero and Negative Rules

Properties of the Function y = x cos(x)

The function y = x cos(x) oscillates with increasing amplitude as x increases, crossing the x-axis multiple times. Understanding its behavior between 3π/2 and 5π/2, including the sign of the function, is crucial for correctly setting up the integral to find the enclosed area.
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Properties of Functions
Related Practice
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 2 of sin(x + 1) dx

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 1 to 2 of x dx

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.

c. u = arctan x

What is the value of the integral?

1
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Textbook Question

Centroid:

Find the centroid of the region cut from the first quadrant by the curve

y = 1/√(x + 1) and the line x = 3.

1
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Textbook Question

Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.

b. Find the centroid of the region.

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 2 to 4 of 1/(s - 1)² ds