Sketch the region enclosed by the curves y = 6 - ² and y = x² - 1. List the points of intersection of these curves from LEFT to RIGHT in the form (x, y): and 0. Decide whether to integrate with respect to x or y to find the area A of this region. A =
Sketch the region enclosed by the curves y = 6 - ² and y = x² - 1. List the points of intersection of these curves from LEFT to RIGHT in the form (x, y): and 0. Decide whether to integrate with respect to x or y to find the area A of this region. A =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Title: Solving for the Area Enclosed by Two Curves**
**Objective:**
Learn how to find the area enclosed by two curves through intersection points and integration.
**Problem Statement:**
Sketch the region enclosed by the curves \( y = 6 - x^2 \) and \( y = x^2 - 1 \).
**Instructions:**
1. **Find Intersection Points:**
List the points of intersection of these curves from LEFT to RIGHT in the form \((x, y)\):
\[
\left( \ \ \ , \ \ \ \right) \ \text{and} \ \left( \ \ \ , \ \ \ \right)
\]
(Note: You need to solve the equations \( 6 - x^2 = x^2 - 1 \) to find the intersection points.)
2. **Determine Integration Boundaries:**
Decide whether to integrate with respect to \( x \) or \( y \) to find the area \( A \) of this region.
\[
A = \ \text{[Fill in the appropriate integral expression to solve for the area]}
\]
**Step-by-Step Solution:**
1. **Sketching the Curves:**
To sketch the curves, plot the equations \( y = 6 - x^2 \) and \( y = x^2 - 1 \). These represent a downward-opening parabola and an upward-opening parabola, respectively.
2. **Finding Intersection Points:**
- Set the equations equal to each other to find the points of intersection:
\[
6 - x^2 = x^2 - 1
\]
- Solve for \( x \):
\[
6 + 1 = 2x^2
\]
\[
7 = 2x^2
\]
\[
x^2 = \frac{7}{2}
\]
\[
x = \pm \sqrt{\frac{7}{2}}
\]
- Substitute \( x \)-values back to find \( y \)-values, yielding the points of intersection \((x, y)\).
3. **Integrate to Find Area:**
- Determine the top and bottom curves from the intersections.
- Integrate the difference between the top and bottom functions with respect](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8fabc15-e5db-480c-94d9-e42b67b656a8%2F53877c0e-7ace-4eaf-8dca-8e47dbd94910%2Fpy4z1hi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Solving for the Area Enclosed by Two Curves**
**Objective:**
Learn how to find the area enclosed by two curves through intersection points and integration.
**Problem Statement:**
Sketch the region enclosed by the curves \( y = 6 - x^2 \) and \( y = x^2 - 1 \).
**Instructions:**
1. **Find Intersection Points:**
List the points of intersection of these curves from LEFT to RIGHT in the form \((x, y)\):
\[
\left( \ \ \ , \ \ \ \right) \ \text{and} \ \left( \ \ \ , \ \ \ \right)
\]
(Note: You need to solve the equations \( 6 - x^2 = x^2 - 1 \) to find the intersection points.)
2. **Determine Integration Boundaries:**
Decide whether to integrate with respect to \( x \) or \( y \) to find the area \( A \) of this region.
\[
A = \ \text{[Fill in the appropriate integral expression to solve for the area]}
\]
**Step-by-Step Solution:**
1. **Sketching the Curves:**
To sketch the curves, plot the equations \( y = 6 - x^2 \) and \( y = x^2 - 1 \). These represent a downward-opening parabola and an upward-opening parabola, respectively.
2. **Finding Intersection Points:**
- Set the equations equal to each other to find the points of intersection:
\[
6 - x^2 = x^2 - 1
\]
- Solve for \( x \):
\[
6 + 1 = 2x^2
\]
\[
7 = 2x^2
\]
\[
x^2 = \frac{7}{2}
\]
\[
x = \pm \sqrt{\frac{7}{2}}
\]
- Substitute \( x \)-values back to find \( y \)-values, yielding the points of intersection \((x, y)\).
3. **Integrate to Find Area:**
- Determine the top and bottom curves from the intersections.
- Integrate the difference between the top and bottom functions with respect
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