Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Integral Evaluation Exercise
Evaluate the integral:
\[ \int_{{-\pi/4}}^{{3\pi/4}} 2 \sec \theta \tan \theta \, d\theta \]
#### Multiple Choice Options:
- **A.** \( 2\sqrt{2} \)
- **B.** \(- 4\sqrt{2} \)
- **C.** \(- 2\sqrt{2} \)
- **D.** \( 0 \)
#### Solution:
First, recognize that the integrand \( 2 \sec \theta \tan \theta \) can be simplified using integration properties. The function \( \sec \theta \tan \theta \) is the derivative of \( \sec \theta \). Therefore, we can rewrite the integral as:
\[ \int_{{-\pi/4}}^{{3\pi/4}} 2 \sec \theta \tan \theta \, d\theta = 2 \int_{{-\pi/4}}^{{3\pi/4}} \sec \theta \tan \theta \, d\theta \]
Let \( u = \sec \theta \):
\[ du = \sec \theta \tan \theta \, d\theta \]
Thus, the integral becomes:
\[ 2 \int_{{-\pi/4}}^{{3\pi/4}} du = 2 [\sec \theta]_{{-\pi/4}}^{{3\pi/4}} \]
Evaluate the bounds:
\[ \sec(-\pi/4) = \sec(\pi/4) = \sqrt{2} \]
\[ \sec(3\pi/4) = -\sqrt{2} \]
Hence:
\[ 2[\sec(3\pi/4) - \sec(-\pi/4)] = 2[-\sqrt{2} - \sqrt{2}] = 2 \times -2\sqrt{2} = -4\sqrt{2} \]
Therefore, the correct answer is:
- **B.** \(- 4\sqrt{2} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fe208ab-ed49-4b20-9691-a6c8abdc4179%2F31969ea5-71b0-4747-a102-5d0cefd55366%2Fo8pwvbb_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Integral Evaluation Exercise
Evaluate the integral:
\[ \int_{{-\pi/4}}^{{3\pi/4}} 2 \sec \theta \tan \theta \, d\theta \]
#### Multiple Choice Options:
- **A.** \( 2\sqrt{2} \)
- **B.** \(- 4\sqrt{2} \)
- **C.** \(- 2\sqrt{2} \)
- **D.** \( 0 \)
#### Solution:
First, recognize that the integrand \( 2 \sec \theta \tan \theta \) can be simplified using integration properties. The function \( \sec \theta \tan \theta \) is the derivative of \( \sec \theta \). Therefore, we can rewrite the integral as:
\[ \int_{{-\pi/4}}^{{3\pi/4}} 2 \sec \theta \tan \theta \, d\theta = 2 \int_{{-\pi/4}}^{{3\pi/4}} \sec \theta \tan \theta \, d\theta \]
Let \( u = \sec \theta \):
\[ du = \sec \theta \tan \theta \, d\theta \]
Thus, the integral becomes:
\[ 2 \int_{{-\pi/4}}^{{3\pi/4}} du = 2 [\sec \theta]_{{-\pi/4}}^{{3\pi/4}} \]
Evaluate the bounds:
\[ \sec(-\pi/4) = \sec(\pi/4) = \sqrt{2} \]
\[ \sec(3\pi/4) = -\sqrt{2} \]
Hence:
\[ 2[\sec(3\pi/4) - \sec(-\pi/4)] = 2[-\sqrt{2} - \sqrt{2}] = 2 \times -2\sqrt{2} = -4\sqrt{2} \]
Therefore, the correct answer is:
- **B.** \(- 4\sqrt{2} \)
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