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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![## Finding the Area of a Region between Two Lines
### Problem Statement
Find the area of the region bounded to the right by \( x = 4 - y \) and to the left by \( x = y \).
### Explanation of the Graph
The accompanying graph is a Cartesian plane with the \( x \)-axis and \( y \)-axis clearly marked. Two intersecting lines are plotted to find the area of their bounded region:
1. **Line 1:** \( x = 4 - y \)
- This line is indicated in green and has a negative slope. It intercepts the y-axis at (0, 4) and the x-axis at (4, 0).
2. **Line 2:** \( x = y \)
- This line is represented in green and has a slope of 1. It passes through the origin (0, 0) and proceeds diagonally, intersecting both the positive x and y-axes at equal intervals.
Below are the key steps required to determine the area of the bounded region:
1. **Intersection Points:**
- To find the intersections, set \( 4 - y = y \).
- Solving the equation \( 4 = 2y \) yields \( y = 2 \).
- Substitute \( y = 2 \) in \( x = y \) to get \( x = 2 \).
- Hence, the intersection points are at (2, 2).
2. **Shaded Region:**
- The shaded area on the graph (indicated via hatching) is bounded by the lines \( x = 4 - y \) and \( x = y \).
### Calculating the Area
Using integration, the area \( A \) can be calculated as follows:
\[ A = \int_{y=0}^{y=2} \left[(4 - y) - y \right] \, dy \]
\[ = \int_{y=0}^{y=2} (4 - 2y) \, dy \]
\[ = \left[4y - y^2\right]_{0}^{2} \]
\[ = \left[(4 \cdot 2 - 2^2) - (4 \cdot 0 - 0^2)\right] \]
\[ = \left[8 - 4\right] = 4 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bc5b610-4c42-40dc-b81a-6ff9e1fb2857%2F9ac4789d-cb2f-45a4-8784-c9c3d916d0d7%2Feqtogwd_processed.png&w=3840&q=75)
Transcribed Image Text:## Finding the Area of a Region between Two Lines
### Problem Statement
Find the area of the region bounded to the right by \( x = 4 - y \) and to the left by \( x = y \).
### Explanation of the Graph
The accompanying graph is a Cartesian plane with the \( x \)-axis and \( y \)-axis clearly marked. Two intersecting lines are plotted to find the area of their bounded region:
1. **Line 1:** \( x = 4 - y \)
- This line is indicated in green and has a negative slope. It intercepts the y-axis at (0, 4) and the x-axis at (4, 0).
2. **Line 2:** \( x = y \)
- This line is represented in green and has a slope of 1. It passes through the origin (0, 0) and proceeds diagonally, intersecting both the positive x and y-axes at equal intervals.
Below are the key steps required to determine the area of the bounded region:
1. **Intersection Points:**
- To find the intersections, set \( 4 - y = y \).
- Solving the equation \( 4 = 2y \) yields \( y = 2 \).
- Substitute \( y = 2 \) in \( x = y \) to get \( x = 2 \).
- Hence, the intersection points are at (2, 2).
2. **Shaded Region:**
- The shaded area on the graph (indicated via hatching) is bounded by the lines \( x = 4 - y \) and \( x = y \).
### Calculating the Area
Using integration, the area \( A \) can be calculated as follows:
\[ A = \int_{y=0}^{y=2} \left[(4 - y) - y \right] \, dy \]
\[ = \int_{y=0}^{y=2} (4 - 2y) \, dy \]
\[ = \left[4y - y^2\right]_{0}^{2} \]
\[ = \left[(4 \cdot 2 - 2^2) - (4 \cdot 0 - 0^2)\right] \]
\[ = \left[8 - 4\right] = 4 \]
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