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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Question
![**Finding the Area Between Curves**
**Problem Statement:**
Find the area of the region bounded above by \( y = x + 2 \) and below by \( y = x^2 \).
**Graphical Representation:**
The image contains a graph that shows:
- The line \( y = x + 2 \) intersecting the parabola \( y = x^2 \).
- The x-axis and y-axis, with numerical labels ranging from -4 to 4.
The graph visually shows the region that is bounded above by the line \( y = x + 2 \) and below by the curve \( y = x^2 \).
**Solution Options:**
1. \( \frac{13}{3} \)
2. \( \frac{9}{2} \) (highlighted as the correct answer)
3. \( \frac{32}{3} \)
4. \( \frac{64}{3} \)
**Detailed Steps to Solve the Problem:**
1. **Find Points of Intersection:**
Solve \( x + 2 = x^2 \):
\[
x^2 - x - 2 = 0
\]
Factoring:
\[
(x - 2)(x + 1) = 0
\]
Thus, \( x = 2 \) and \( x = -1 \).
2. **Set Up the Integral:**
The area \( A \) between the two curves from \( x = -1 \) to \( x = 2 \) is given by:
\[
A = \int_{-1}^{2} [(x + 2) - x^2] \, dx
\]
3. **Calculate the Integral:**
\[
\int_{-1}^{2} (x + 2 - x^2) \, dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}
\]
Substitute the bounds into the antiderivative:
\[
\left(\frac{2^2}{2} + 2(2) - \frac{2^3}{3}\right) - \left(\frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb111f9a2-d607-44d0-9cf1-983102308b09%2F8e3635cc-c54a-48c4-80f8-f50d68bc73af%2Fo9fq81m_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding the Area Between Curves**
**Problem Statement:**
Find the area of the region bounded above by \( y = x + 2 \) and below by \( y = x^2 \).
**Graphical Representation:**
The image contains a graph that shows:
- The line \( y = x + 2 \) intersecting the parabola \( y = x^2 \).
- The x-axis and y-axis, with numerical labels ranging from -4 to 4.
The graph visually shows the region that is bounded above by the line \( y = x + 2 \) and below by the curve \( y = x^2 \).
**Solution Options:**
1. \( \frac{13}{3} \)
2. \( \frac{9}{2} \) (highlighted as the correct answer)
3. \( \frac{32}{3} \)
4. \( \frac{64}{3} \)
**Detailed Steps to Solve the Problem:**
1. **Find Points of Intersection:**
Solve \( x + 2 = x^2 \):
\[
x^2 - x - 2 = 0
\]
Factoring:
\[
(x - 2)(x + 1) = 0
\]
Thus, \( x = 2 \) and \( x = -1 \).
2. **Set Up the Integral:**
The area \( A \) between the two curves from \( x = -1 \) to \( x = 2 \) is given by:
\[
A = \int_{-1}^{2} [(x + 2) - x^2] \, dx
\]
3. **Calculate the Integral:**
\[
\int_{-1}^{2} (x + 2 - x^2) \, dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}
\]
Substitute the bounds into the antiderivative:
\[
\left(\frac{2^2}{2} + 2(2) - \frac{2^3}{3}\right) - \left(\frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^
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