B Find f (x + x-¹) dx B5 +In(B) - -33234 5 15 32 B5 + In(B) − (³² + ln(2)) 5 B5 + ln(B) − ln(2) - where B > 2 32 B5 + In(B) - (3² + In (2)) 5 B5+ln(B) - 0
B Find f (x + x-¹) dx B5 +In(B) - -33234 5 15 32 B5 + In(B) − (³² + ln(2)) 5 B5 + ln(B) − ln(2) - where B > 2 32 B5 + In(B) - (3² + In (2)) 5 B5+ln(B) - 0
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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![### Calculating an Integral for Educational Purposes
In this exercise, we are asked to find the integral of the function \(x^4 + x^{-1}\) from 2 to \(B\), where \( B > 2 \). The problem is given as:
\[ \int_{2}^{B} (x^4 + x^{-1}) \, dx \]
Here are the steps and solution options for this problem:
1. **Integral Definition:**
The given integral is:
\[ \int_{2}^{B} (x^4 + x^{-1}) \, dx \]
2. **Integration:**
To solve this integral, we need to integrate each term separately:
\[ \int x^4 \, dx + \int x^{-1} \, dx \]
- The integral of \(x^4\) is:
\[ \int x^4 \, dx = \frac{1}{5} x^5 \]
- The integral of \(x^{-1}\) is:
\[ \int x^{-1} \, dx = \ln(x) \]
3. **Combining the Results:**
By combining the integrals, we get:
\[ \left[ \frac{1}{5} x^5 + \ln(x) \right]_2^B \]
4. **Evaluating the Definite Integral:**
Substitute the upper and lower limits into the result:
\[ \left( \frac{1}{5} B^5 + \ln(B) \right) - \left( \frac{1}{5} (2)^5 + \ln(2) \right) \]
5. **Simplifying the Expression:**
Simplify the expression:
\[ \frac{1}{5} B^5 + \ln(B) - \left( \frac{32}{5} + \ln(2) \right) \]
### Multiple Choice Options:
The options provided are:
- \( \frac{1}{5} B^5 + \ln(B) - \left( \frac{32}{5} + \ln(2) \right) \)
- \( \frac{1}{5} B^5 + \ln(B) - \frac{32}{5} \)
- \( \frac{1}{5} B^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F926c3fce-ea0b-442b-8045-2d4039975256%2F358e89b6-f61e-47fe-8b8b-2e3e4a1cfc11%2Fryhz6m_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating an Integral for Educational Purposes
In this exercise, we are asked to find the integral of the function \(x^4 + x^{-1}\) from 2 to \(B\), where \( B > 2 \). The problem is given as:
\[ \int_{2}^{B} (x^4 + x^{-1}) \, dx \]
Here are the steps and solution options for this problem:
1. **Integral Definition:**
The given integral is:
\[ \int_{2}^{B} (x^4 + x^{-1}) \, dx \]
2. **Integration:**
To solve this integral, we need to integrate each term separately:
\[ \int x^4 \, dx + \int x^{-1} \, dx \]
- The integral of \(x^4\) is:
\[ \int x^4 \, dx = \frac{1}{5} x^5 \]
- The integral of \(x^{-1}\) is:
\[ \int x^{-1} \, dx = \ln(x) \]
3. **Combining the Results:**
By combining the integrals, we get:
\[ \left[ \frac{1}{5} x^5 + \ln(x) \right]_2^B \]
4. **Evaluating the Definite Integral:**
Substitute the upper and lower limits into the result:
\[ \left( \frac{1}{5} B^5 + \ln(B) \right) - \left( \frac{1}{5} (2)^5 + \ln(2) \right) \]
5. **Simplifying the Expression:**
Simplify the expression:
\[ \frac{1}{5} B^5 + \ln(B) - \left( \frac{32}{5} + \ln(2) \right) \]
### Multiple Choice Options:
The options provided are:
- \( \frac{1}{5} B^5 + \ln(B) - \left( \frac{32}{5} + \ln(2) \right) \)
- \( \frac{1}{5} B^5 + \ln(B) - \frac{32}{5} \)
- \( \frac{1}{5} B^
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