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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.R.122
Chapter 8, Problem 8.R.122

122. Comparing areas The region R₁ is bounded by the graph of y = tan(x) and the x-axis on the interval [0, π/3].
The region R₂ is bounded by the graph of y = sec(x) and the x-axis on the interval [0, π/6]. Which region has the greater area?

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Identify the two regions and their boundaries: Region R₁ is bounded by y = tan(x) and the x-axis on [0, \(\frac{\pi}{3}\)], and Region R₂ is bounded by y = sec(x) and the x-axis on [0, \(\frac{\pi}{6}\)].
Set up the definite integrals to find the areas of each region. The area of R₁ is given by the integral \( A_1 = \int_0^{\frac{\pi}{3}} \tan(x) \, dx \), and the area of R₂ is given by \( A_2 = \int_0^{\frac{\pi}{6}} \sec(x) \, dx \).
Recall the antiderivatives needed: The integral of \( \tan(x) \) is \( -\ln|\cos(x)| + C \), and the integral of \( \sec(x) \) is \( \ln|\sec(x) + \tan(x)| + C \).
Evaluate each definite integral by applying the Fundamental Theorem of Calculus, substituting the upper and lower limits into the antiderivatives for both \( A_1 \) and \( A_2 \).
Compare the two computed areas \( A_1 \) and \( A_2 \) to determine which region has the greater area.

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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 graph of the function and the x-axis. For positive functions, this integral gives the exact area of the region bounded by the curve and the axis. Calculating these integrals helps compare areas of different regions.
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Definition of the Definite Integral

Properties of the Functions tan(x) and sec(x)

The functions tan(x) and sec(x) are trigonometric functions with distinct behaviors on given intervals. Understanding their values and growth on [0, π/3] and [0, π/6] respectively is essential, as tan(x) increases from 0 to √3, while sec(x) increases from 1 to 2, affecting the area under their curves.
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Properties of Functions

Comparison of Areas via Integration Limits and Function Behavior

Comparing areas requires evaluating integrals over different intervals and functions. Since the intervals and functions differ, analyzing how the length of the interval and the magnitude of the function values influence the total area is crucial to determine which region has the greater area.
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Limit Comparison Test
Related Practice
Textbook Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

25. ∫ (from -3/2 to -1) dx/(4x² + 12x + 10)

Textbook Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

63. ∫ dx/(x² - 2x - 15)

1
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Textbook Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

6. ∫ (2 − sin 2θ)/cos² 2θ dθ

Textbook Question

118. Two worthy integrals

b. Let f be any positive continuous function on the interval [0, π/2]. Evaluate

∫ from 0 to π/2 of [f(cos x) / (f(cos x) + f(sin x))] dx.

(Hint: Use the identity cos(π/2 − x) = sin x.)


(Source: Mathematics Magazine 81, 2, Apr 2008)

Textbook Question

120. Equal volumes

a. Let R be the region bounded by the graph of f(x) = x^(-p) and the x-axis, for x ≥ 1. Let V₁ and V₂ be the volumes of the solids generated when R is revolved about the x-axis and the y-axis, respectively, if they exist. For what values of p (if any) is V₁ = V₂?

b. Repeat part (a) on the interval [0, 1].

Textbook Question

119. {Use of Tech} Comparing volumes Let R be the region bounded by y = ln(x), the x-axis, and the line x = a, where a > 1.

b. Find the volume V₂(a) of the solid generated when R is revolved about the y-axis (as a function of a).