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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Question
![**Problem Statement:**
Use the trigonometric substitution \( x = 2 \sin \theta \) to compute the integral
\[
\int_{-2}^{2} \sqrt{4 - x^2} \, dx
\]
**Solution Explanation:**
This problem involves using a trigonometric substitution to evaluate the integral of a square root function. The substitution \( x = 2 \sin \theta \) is typical for integrals involving square roots of the form \( \sqrt{a^2 - x^2} \), which in this case is \( \sqrt{4 - x^2} \).
**Steps:**
1. **Substitute** \( x = 2 \sin \theta \), which implies \( dx = 2 \cos \theta \, d\theta \).
2. **Change Limits of Integration:**
- When \( x = -2 \), \( \sin \theta = -1 \), so \( \theta = -\frac{\pi}{2} \).
- When \( x = 2 \), \( \sin \theta = 1 \), so \( \theta = \frac{\pi}{2} \).
3. **Express the Integral in terms of \(\theta\):**
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{4 - (2 \sin \theta)^2} \cdot 2 \cos \theta \, d\theta
\]
Simplify inside the square root:
\[
\sqrt{4 - 4 \sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta
\]
4. **Simplified Integral:**
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2 \cos \theta \cdot 2 \cos \theta \, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4 \cos^2 \theta \, d\theta
\]
5. **Evaluate the Integral:**
Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ec25ffb-8aef-468b-ad53-a5caf8bbc7e2%2Fdd5c5989-664a-451e-80aa-00f672facc40%2F8w6wa2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Use the trigonometric substitution \( x = 2 \sin \theta \) to compute the integral
\[
\int_{-2}^{2} \sqrt{4 - x^2} \, dx
\]
**Solution Explanation:**
This problem involves using a trigonometric substitution to evaluate the integral of a square root function. The substitution \( x = 2 \sin \theta \) is typical for integrals involving square roots of the form \( \sqrt{a^2 - x^2} \), which in this case is \( \sqrt{4 - x^2} \).
**Steps:**
1. **Substitute** \( x = 2 \sin \theta \), which implies \( dx = 2 \cos \theta \, d\theta \).
2. **Change Limits of Integration:**
- When \( x = -2 \), \( \sin \theta = -1 \), so \( \theta = -\frac{\pi}{2} \).
- When \( x = 2 \), \( \sin \theta = 1 \), so \( \theta = \frac{\pi}{2} \).
3. **Express the Integral in terms of \(\theta\):**
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{4 - (2 \sin \theta)^2} \cdot 2 \cos \theta \, d\theta
\]
Simplify inside the square root:
\[
\sqrt{4 - 4 \sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta
\]
4. **Simplified Integral:**
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2 \cos \theta \cdot 2 \cos \theta \, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4 \cos^2 \theta \, d\theta
\]
5. **Evaluate the Integral:**
Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\
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