The velocity of a particle on the time interval 1 ≤ x ≤ 4 is v(x) = e²x, where v(x) is in cm/s. What is the average velocity for the interval? O 375.645 396.426 O 475.536 O 495.595 T
The velocity of a particle on the time interval 1 ≤ x ≤ 4 is v(x) = e²x, where v(x) is in cm/s. What is the average velocity for the interval? O 375.645 396.426 O 475.536 O 495.595 T
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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![### Average Velocity Calculation
#### Problem Statement:
The velocity of a particle on the time interval \(1 \leq x \leq 4\) is \( v(x) = e^{2x} \), where \( v(x) \) is in cm/s.
**Question:**
What is the average velocity for the interval?
#### Options:
```
○ 375.645
○ 396.426
○ 475.536
○ 495.595
```
#### Explanation:
To find the average velocity of the particle over the given interval, we use the formula for average value of a continuous function \(v(x)\) over the interval \([a, b]\):
\[ \text{Average velocity} = \frac{1}{b - a} \int_{a}^{b} v(x) \, dx \]
For this problem:
- \( a = 1 \)
- \( b = 4 \)
- \( v(x) = e^{2x} \)
Therefore, we need to calculate the following integral:
\[ \text{Average velocity} = \frac{1}{4 - 1} \int_{1}^{4} e^{2x} \, dx = \frac{1}{3} \int_{1}^{4} e^{2x} \, dx \]
Calculating the integral:
1. \[ \int e^{2x} \, dx \]
Substitute \(u = 2x\), then \(du = 2dx\), and \(dx = \frac{du}{2}\):
\[ \int e^{2x} \, dx = \frac{1}{2} \int e^{u} \, du = \frac{1}{2} e^{u}\]
Substitute back \(u = 2x\):
\[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \]
2. Evaluate the definite integral \(\int_{1}^{4} e^{2x} \, dx\):
\[ \int_{1}^{4} e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_{1}^{4} = \frac{1}{2} (e^{8} - e^{2}) \]
3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fb4dfc3-db65-4bb6-9a5b-9427f46c7a3d%2F08708ba9-32c2-4ee1-a233-b5c7da8ae596%2Feodydg_processed.png&w=3840&q=75)
Transcribed Image Text:### Average Velocity Calculation
#### Problem Statement:
The velocity of a particle on the time interval \(1 \leq x \leq 4\) is \( v(x) = e^{2x} \), where \( v(x) \) is in cm/s.
**Question:**
What is the average velocity for the interval?
#### Options:
```
○ 375.645
○ 396.426
○ 475.536
○ 495.595
```
#### Explanation:
To find the average velocity of the particle over the given interval, we use the formula for average value of a continuous function \(v(x)\) over the interval \([a, b]\):
\[ \text{Average velocity} = \frac{1}{b - a} \int_{a}^{b} v(x) \, dx \]
For this problem:
- \( a = 1 \)
- \( b = 4 \)
- \( v(x) = e^{2x} \)
Therefore, we need to calculate the following integral:
\[ \text{Average velocity} = \frac{1}{4 - 1} \int_{1}^{4} e^{2x} \, dx = \frac{1}{3} \int_{1}^{4} e^{2x} \, dx \]
Calculating the integral:
1. \[ \int e^{2x} \, dx \]
Substitute \(u = 2x\), then \(du = 2dx\), and \(dx = \frac{du}{2}\):
\[ \int e^{2x} \, dx = \frac{1}{2} \int e^{u} \, du = \frac{1}{2} e^{u}\]
Substitute back \(u = 2x\):
\[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \]
2. Evaluate the definite integral \(\int_{1}^{4} e^{2x} \, dx\):
\[ \int_{1}^{4} e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_{1}^{4} = \frac{1}{2} (e^{8} - e^{2}) \]
3.
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