1 Si 1+z2 da is convergent because 1+z² Jhry sin x and ₁ sin² x dx is convergent. sin² 2.₁ side is convergent because ≤ 1 and 2 do is convergent. sin² x 1+x2 1+2 a 3. So sin da is convergent because in and da is convergent. < 4. Sin da is divergent because in and fd is divergent. 21 1+x² 1+x²
1 Si 1+z2 da is convergent because 1+z² Jhry sin x and ₁ sin² x dx is convergent. sin² 2.₁ side is convergent because ≤ 1 and 2 do is convergent. sin² x 1+x2 1+2 a 3. So sin da is convergent because in and da is convergent. < 4. Sin da is divergent because in and fd is divergent. 21 1+x² 1+x²
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 Convergence Assessment
In this exercise, the convergence of the improper integral
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx
\]
is analyzed under different conditions to determine whether it converges or diverges.
1. **Option 1**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is convergent because } \frac{\sin^2{x}}{1+x^2} \leq \sin^2{x} \text{ and } \int_{1}^{\infty} \sin^2{x} \, dx \text{ is convergent.}
\]
2. **Option 2**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is convergent because } \frac{\sin^2{x}}{1+x^2} \leq \frac{1}{1+x^2} \text{ and } \int_{1}^{\infty} \frac{1}{1+x^2} \, dx \text{ is convergent.}
\]
3. **Option 3**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is convergent because } \frac{\sin^2{x}}{1+x^2} \leq \frac{1}{x^2} \text{ and } \int_{1}^{\infty} \frac{1}{x^2} \, dx \text{ is convergent.}
\]
4. **Option 4**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is divergent because } \frac{\sin^2{x}}{1+x^2} \geq \frac{1}{1+x^2} \text{ and } \int_{1}^{\infty} \frac{1}{1+x^2} \, dx \text{ is divergent.}
\]
**Analysis:**
- To determine the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8fabc15-e5db-480c-94d9-e42b67b656a8%2Fec6b8d59-38e6-4c26-b46f-4ffdde3013c6%2F9n97h5q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Integral Convergence Assessment
In this exercise, the convergence of the improper integral
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx
\]
is analyzed under different conditions to determine whether it converges or diverges.
1. **Option 1**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is convergent because } \frac{\sin^2{x}}{1+x^2} \leq \sin^2{x} \text{ and } \int_{1}^{\infty} \sin^2{x} \, dx \text{ is convergent.}
\]
2. **Option 2**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is convergent because } \frac{\sin^2{x}}{1+x^2} \leq \frac{1}{1+x^2} \text{ and } \int_{1}^{\infty} \frac{1}{1+x^2} \, dx \text{ is convergent.}
\]
3. **Option 3**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is convergent because } \frac{\sin^2{x}}{1+x^2} \leq \frac{1}{x^2} \text{ and } \int_{1}^{\infty} \frac{1}{x^2} \, dx \text{ is convergent.}
\]
4. **Option 4**
\[
\int_{1}^{\infty} \frac{\sin^2{x}}{1+x^2} \, dx \text{ is divergent because } \frac{\sin^2{x}}{1+x^2} \geq \frac{1}{1+x^2} \text{ and } \int_{1}^{\infty} \frac{1}{1+x^2} \, dx \text{ is divergent.}
\]
**Analysis:**
- To determine the
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