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 Bounded Region
**Problem Statement:**
Find the area of the region bounded to the right by \( x = y + 2 \) and to the left by \( x = y^2 \).
**Graphical Representation:**
The provided graph plots two functions:
1. The line \( x = y + 2 \) (a linear function).
2. The parabola \( x = y^2 \) (a quadratic function).
**Graph Explanation:**
- **Axes:**
- The horizontal axis represents the \( x \)-axis.
- The vertical axis represents the \( y \)-axis.
- Both axes range from -5 to 5 in increments of 1.
- **Function Plots:**
- The green curve representing the quadratic function \( x = y^2 \) opens to the right, intersecting the \( y \)-axis at the origin (0,0).
- The green line representing the linear function \( x = y + 2 \) intersects the \( y \)-axis at (2,0) and rises with a slope of 1.
**Intersection Points:**
To find the bounded region's area, we need to determine the intersection points of the functions \( x = y + 2 \) and \( x = y^2 \).
**Setting the equations equal:**
\[ y + 2 = y^2 \]
Solving for \( y \):
\[ y^2 - y - 2 = 0 \]
\[ (y - 2)(y + 1) = 0 \]
Therefore, the intersections occur at:
\[ y = 2 \quad \text{and} \quad y = -1 \]
**Setting Up the Integral:**
To find the area between these curves, we integrate with respect to \( y \):
\[ \text{Area} = \int_{-1}^{2} ( (y + 2) - y^2 ) \, dy \]
Integrate term-by-term:
\[ \int_{-1}^{2} (y + 2 - y^2) \, dy \]
\[ = \int_{-1}^{2} y \, dy + \int_{-1}^{2} 2 \, dy - \int_{-1}^{2} y^2 \, dy \]
Calculating these integrals individually:
\[ \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bc5b610-4c42-40dc-b81a-6ff9e1fb2857%2Fe450d14e-dd03-4ea7-96ad-b970cbf91c5b%2Fange5il_processed.png&w=3840&q=75)
Transcribed Image Text:### Finding the Area of a Bounded Region
**Problem Statement:**
Find the area of the region bounded to the right by \( x = y + 2 \) and to the left by \( x = y^2 \).
**Graphical Representation:**
The provided graph plots two functions:
1. The line \( x = y + 2 \) (a linear function).
2. The parabola \( x = y^2 \) (a quadratic function).
**Graph Explanation:**
- **Axes:**
- The horizontal axis represents the \( x \)-axis.
- The vertical axis represents the \( y \)-axis.
- Both axes range from -5 to 5 in increments of 1.
- **Function Plots:**
- The green curve representing the quadratic function \( x = y^2 \) opens to the right, intersecting the \( y \)-axis at the origin (0,0).
- The green line representing the linear function \( x = y + 2 \) intersects the \( y \)-axis at (2,0) and rises with a slope of 1.
**Intersection Points:**
To find the bounded region's area, we need to determine the intersection points of the functions \( x = y + 2 \) and \( x = y^2 \).
**Setting the equations equal:**
\[ y + 2 = y^2 \]
Solving for \( y \):
\[ y^2 - y - 2 = 0 \]
\[ (y - 2)(y + 1) = 0 \]
Therefore, the intersections occur at:
\[ y = 2 \quad \text{and} \quad y = -1 \]
**Setting Up the Integral:**
To find the area between these curves, we integrate with respect to \( y \):
\[ \text{Area} = \int_{-1}^{2} ( (y + 2) - y^2 ) \, dy \]
Integrate term-by-term:
\[ \int_{-1}^{2} (y + 2 - y^2) \, dy \]
\[ = \int_{-1}^{2} y \, dy + \int_{-1}^{2} 2 \, dy - \int_{-1}^{2} y^2 \, dy \]
Calculating these integrals individually:
\[ \
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