Textbook Question
The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀¹ dr / r^0.999
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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 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 8.3.78
Average Value: Find the average value of the function f(x) = 1 / (1 - sin θ) on the interval [0, π/6].
Verified step by step guidance1
Recall the formula for the average value of a function \( f(x) \) on the interval \( [a, b] \):
\[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx \]
Identify the function and interval given: \( f(\theta) = \frac{1}{1 - \sin \theta} \), with \( a = 0 \) and \( b = \frac{\pi}{6} \).
Set up the integral for the average value:
\[ \frac{1}{\frac{\pi}{6} - 0} \int_0^{\frac{\pi}{6}} \frac{1}{1 - \sin \theta} \, d\theta = \frac{6}{\pi} \int_0^{\frac{\pi}{6}} \frac{1}{1 - \sin \theta} \, d\theta \]
To evaluate the integral \( \int \frac{1}{1 - \sin \theta} \, d\theta \), use a trigonometric identity or substitution. One common approach is to multiply numerator and denominator by the conjugate \( 1 + \sin \theta \) to simplify the integrand.
After simplifying, integrate the resulting expression with respect to \( \theta \) from 0 to \( \frac{\pi}{6} \), then multiply by \( \frac{6}{\pi} \) to find the average value.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Average Value of a Function
The average value of a function f(x) on an interval [a, b] is given by (1/(b - a)) times the definite integral of f(x) from a to b. It represents the mean height of the function over that interval.
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Average Value of a Function
Definite Integral
A definite integral calculates the accumulated area under the curve of a function between two points a and b. It is essential for finding the total accumulation or average value over an interval.
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Integration of Trigonometric Functions
Integrating functions involving trigonometric expressions like 1/(1 - sin θ) often requires substitution or trigonometric identities. Understanding these techniques helps simplify and evaluate the integral accurately.
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Related Practice
Textbook Question
Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.
∫ dx / (x² √(x² + 1))
Textbook Question
Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀^π/2 x³ cos 2x dx
Textbook Question
In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (2x + 1) / (x² - 7x + 12) dx
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Textbook Question
Evaluate the integrals in Exercises 1–22.
∫ cos³(4x) dx
Textbook Question
The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀^∞ dx / [(x + 1)(x² + 1)]
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