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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![**Problem Statement:**
Given the graph to the right, find the integral of \( f(x) \) from 3 to 9.
\[ \int_{3}^{9} f(x) \, dx \]
**Graph Description:**
The graph provided is a piecewise linear function with the following segments:
1. From \( x = 0 \) to \( x = 2 \), \( f(x) \) is a horizontal line with \( f(x) = 5 \).
2. From \( x = 2 \) to \( x = 5 \), \( f(x) \) is a horizontal line with \( f(x) = 4 \).
3. From \( x = 5 \) to \( x = 8 \), \( f(x) \) is a linear line that increases from \( f(x) = 4 \) to \( f(x) = 8 \).
4. From \( x = 8 \) to \( x = 9 \), \( f(x) \) is a horizontal line with \( f(x) = 8 \).
**Solution:**
To solve the integral \( \int_{3}^{9} f(x) \, dx \), we need to find the area under the curve from \( x = 3 \) to \( x = 9 \).
1. **Interval [3, 5]:**
- \( f(x) = 4 \)
- Area = \( \text{width} \times \text{height} = (5 - 3) \times 4 = 2 \times 4 = 8 \)
2. **Interval [5, 8]:**
- \( f(x) \) is a linear line increasing from 4 to 8.
- Area = Area of trapezoid = \( \frac{1}{2} \times (b1 + b2) \times h = \frac{1}{2} \times (4 + 8) \times 3 = \frac{1}{2} \times 12 \times 3 = 18 \)
3. **Interval [8, 9]:**
- \( f(x) = 8 \)
- Area = \( \text{width} \times \text{height} = (9 - 8) \times 8 = 1 \times 8 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8fac4acb-b08c-48a0-98ba-e3d343d1c48b%2F68085217-ee41-411a-9314-113f5dd3a582%2F2n71gu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given the graph to the right, find the integral of \( f(x) \) from 3 to 9.
\[ \int_{3}^{9} f(x) \, dx \]
**Graph Description:**
The graph provided is a piecewise linear function with the following segments:
1. From \( x = 0 \) to \( x = 2 \), \( f(x) \) is a horizontal line with \( f(x) = 5 \).
2. From \( x = 2 \) to \( x = 5 \), \( f(x) \) is a horizontal line with \( f(x) = 4 \).
3. From \( x = 5 \) to \( x = 8 \), \( f(x) \) is a linear line that increases from \( f(x) = 4 \) to \( f(x) = 8 \).
4. From \( x = 8 \) to \( x = 9 \), \( f(x) \) is a horizontal line with \( f(x) = 8 \).
**Solution:**
To solve the integral \( \int_{3}^{9} f(x) \, dx \), we need to find the area under the curve from \( x = 3 \) to \( x = 9 \).
1. **Interval [3, 5]:**
- \( f(x) = 4 \)
- Area = \( \text{width} \times \text{height} = (5 - 3) \times 4 = 2 \times 4 = 8 \)
2. **Interval [5, 8]:**
- \( f(x) \) is a linear line increasing from 4 to 8.
- Area = Area of trapezoid = \( \frac{1}{2} \times (b1 + b2) \times h = \frac{1}{2} \times (4 + 8) \times 3 = \frac{1}{2} \times 12 \times 3 = 18 \)
3. **Interval [8, 9]:**
- \( f(x) = 8 \)
- Area = \( \text{width} \times \text{height} = (9 - 8) \times 8 = 1 \times 8 =
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