Find the area of the region enclosed by the curves y=x²-7 and y=9. The area of the region enclosed by the curves is (Type a simplified fraction.)
Find the area of the region enclosed by the curves y=x²-7 and y=9. The area of the region enclosed by the curves is (Type a simplified fraction.)
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 and Integrating Areas Between Curves**
**Problem Statement:**
Find the area of the region enclosed by the curves \( y = x^2 - 7 \) and \( y = 9 \).
**Calculation Approach:**
To find the area of the region enclosed by the given curves, follow these steps:
1. **Find Intersection Points:**
- Determine the x-values where the curves intersect by setting the equations equal to each other:
\[ x^2 - 7 = 9 \]
\[ x^2 = 16 \]
\[ x = \pm 4 \]
2. **Set Up the Integral:**
- The area can be found by integrating the difference between the top function and the bottom function from the left intersection point to the right intersection point.
- The top function: \( y = 9 \)
- The bottom function: \( y = x^2 - 7 \)
- The integral should be set up as:
\[
\text{Area} = \int_{-4}^{4} [9 - (x^2 - 7)] \, dx
\]
3. **Simplify the Integrand:**
\[
\text{Area} = \int_{-4}^{4} [9 - x^2 + 7] \, dx
\]
\[
\text{Area} = \int_{-4}^{4} [16 - x^2] \, dx
\]
4. **Evaluate the Integral:**
- Split the integral and solve:
\[
\text{Area} = \int_{-4}^{4} 16 \, dx - \int_{-4}^{4} x^2 \, dx
\]
- For \(\int_{-4}^{4} 16 \, dx\):
\[
16 \times (4 - (-4)) = 16 \times 8 = 128
\]
- For \(\int_{-4}^{4} x^2 \, dx\), note the symmetry property (the integral of an odd function over symmetric limits [-a, a] is zero). Therefore:
\[
\int_{-4}^{4} x^2 \, dx = 2 \int_{0}^{4} x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c5e517d-57db-4b65-baa5-8a292d97b7f5%2F32bb638a-12c4-42ac-a8af-c4ad020f9092%2Fcc9ozy_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding and Integrating Areas Between Curves**
**Problem Statement:**
Find the area of the region enclosed by the curves \( y = x^2 - 7 \) and \( y = 9 \).
**Calculation Approach:**
To find the area of the region enclosed by the given curves, follow these steps:
1. **Find Intersection Points:**
- Determine the x-values where the curves intersect by setting the equations equal to each other:
\[ x^2 - 7 = 9 \]
\[ x^2 = 16 \]
\[ x = \pm 4 \]
2. **Set Up the Integral:**
- The area can be found by integrating the difference between the top function and the bottom function from the left intersection point to the right intersection point.
- The top function: \( y = 9 \)
- The bottom function: \( y = x^2 - 7 \)
- The integral should be set up as:
\[
\text{Area} = \int_{-4}^{4} [9 - (x^2 - 7)] \, dx
\]
3. **Simplify the Integrand:**
\[
\text{Area} = \int_{-4}^{4} [9 - x^2 + 7] \, dx
\]
\[
\text{Area} = \int_{-4}^{4} [16 - x^2] \, dx
\]
4. **Evaluate the Integral:**
- Split the integral and solve:
\[
\text{Area} = \int_{-4}^{4} 16 \, dx - \int_{-4}^{4} x^2 \, dx
\]
- For \(\int_{-4}^{4} 16 \, dx\):
\[
16 \times (4 - (-4)) = 16 \times 8 = 128
\]
- For \(\int_{-4}^{4} x^2 \, dx\), note the symmetry property (the integral of an odd function over symmetric limits [-a, a] is zero). Therefore:
\[
\int_{-4}^{4} x^2 \, dx = 2 \int_{0}^{4} x
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