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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For this question. I'm confused why the answer is pi/4. I circled in green where I was trying to work it out at the end. I understand partial fraction decomp , just not understanding the end part how we got pi/4. I thought answer was infinity because infinity - a number is infinity.
![∫₀∞ (1/(x + 1)√(x² + 1)) dx = lim (b → ∞) [ (1/2) ln|x + 1| - (1/2) ln(x² + 1) + (1/2) tan⁻¹x ]₀ᵇ
= lim (b → ∞) [ (1/2) ln( (b + 1) / √(b² + 1) ) + (1/2) tan⁻¹b ] - [(1/2) ln(1/√1) + (1/2) tan⁻¹0] = (1/2) ln1 + (1/2) - (1/2) ln1 - (1/2) x 0 = π/4
3. Write out the first five terms of the sequence...](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F60e1de78-324b-4574-bf4c-4017b6f47be4%2Futfxitl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:∫₀∞ (1/(x + 1)√(x² + 1)) dx = lim (b → ∞) [ (1/2) ln|x + 1| - (1/2) ln(x² + 1) + (1/2) tan⁻¹x ]₀ᵇ
= lim (b → ∞) [ (1/2) ln( (b + 1) / √(b² + 1) ) + (1/2) tan⁻¹b ] - [(1/2) ln(1/√1) + (1/2) tan⁻¹0] = (1/2) ln1 + (1/2) - (1/2) ln1 - (1/2) x 0 = π/4
3. Write out the first five terms of the sequence...
![**Transcription of Mathematical Notes for Educational Website**
---
Given problem:
\[ \int \frac{1}{x^2+1} \, dx \]
Substitution method:
1. **Substitution 1:**
- Set \( u = x^2 + 1 \)
- Then, \( du = 2x \, dx \)
2. **Substitution 2:**
- Set \( u = x^2 \)
- Then, \( du = 2x \, dx \)
---
**Integration Steps:**
\[ \int \frac{1}{2} \frac{1}{x^2+1} \, dx = \frac{1}{2} \left(\frac{-1}{x^2+1}\right) \left(-1\right) \]
\[ = \frac{-1}{2} \left( - \frac{1}{x^2+1} \right) \]
\[ = \frac{1}{2} \ln |x^2+1| + \text{const} \]
\[ = \frac{1}{2} \ln|x^2+1| + \tan^{-1}x \]
---
**Evaluating the Limits from \(-\infty\) to \(+\infty\):**
---
(Calculation is within a green boxed region)
1. Limit as \( x \to +\infty \):
\[ \int_{0}^{+\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | +\infty + 1| \]
\[ = \frac{1}{2} \ln \infty = \text{undefined} \]
2. Limit as \( x \to 0 \):
\[ \int_{0}^{0} \frac{1}{x^2+1} \, dx = 0 \]
3. Limit as \( x \to -\infty \):
\[ \int_{0}^{-\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | -\infty + 1 | \]
\[ = \frac{1}{2} \ln \infty = \text{undefined} \]
Computation yields results which suggest careful handling of limits,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F60e1de78-324b-4574-bf4c-4017b6f47be4%2Fk24ieb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription of Mathematical Notes for Educational Website**
---
Given problem:
\[ \int \frac{1}{x^2+1} \, dx \]
Substitution method:
1. **Substitution 1:**
- Set \( u = x^2 + 1 \)
- Then, \( du = 2x \, dx \)
2. **Substitution 2:**
- Set \( u = x^2 \)
- Then, \( du = 2x \, dx \)
---
**Integration Steps:**
\[ \int \frac{1}{2} \frac{1}{x^2+1} \, dx = \frac{1}{2} \left(\frac{-1}{x^2+1}\right) \left(-1\right) \]
\[ = \frac{-1}{2} \left( - \frac{1}{x^2+1} \right) \]
\[ = \frac{1}{2} \ln |x^2+1| + \text{const} \]
\[ = \frac{1}{2} \ln|x^2+1| + \tan^{-1}x \]
---
**Evaluating the Limits from \(-\infty\) to \(+\infty\):**
---
(Calculation is within a green boxed region)
1. Limit as \( x \to +\infty \):
\[ \int_{0}^{+\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | +\infty + 1| \]
\[ = \frac{1}{2} \ln \infty = \text{undefined} \]
2. Limit as \( x \to 0 \):
\[ \int_{0}^{0} \frac{1}{x^2+1} \, dx = 0 \]
3. Limit as \( x \to -\infty \):
\[ \int_{0}^{-\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | -\infty + 1 | \]
\[ = \frac{1}{2} \ln \infty = \text{undefined} \]
Computation yields results which suggest careful handling of limits,
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