Skip to main content
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 8.3.78
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 guidance
1
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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

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.
Recommended video:
06:37
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.
Recommended video:
05:43
Definition of the Definite Integral

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.
Recommended video:
6:04
Introduction to Trigonometric Functions
Related Practice
Textbook Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 0 to 1 of (dt / (t - sin t))

(Hint: t ≥ sin t for t ≥ 0)

1
views
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

1
views
Textbook Question

Evaluate the integrals in Exercises 1–22.

∫ cos³(4x) dx

Textbook Question

Evaluate the integrals in Exercises 33–52.

∫ cot⁶(2x) dx

Textbook Question

In Exercises 67–73, use integration by parts to establish the reduction formula.

∫ x^n sin(x) dx = -x^n cos(x) + n ∫ x^(n-1) cos(x) dx

Textbook Question

Exercises 59–64 require the use of various trigonometric identities before you evaluate the integrals.

∫ cos²(2θ) sin(θ) dθ