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
Please use just calculus II
![### Problem 3: Finding Area Bounded by Curves
#### Instructions:
**Graph and then find the area of the region bounded by the curves \( y = 8x - x^2 \) and \( y = 2x - 7 \).**
#### Steps:
1. **Graph the Curves:**
- The first curve is \( y = 8x - x^2 \), which is a downward-opening parabola.
- The second curve is \( y = 2x - 7 \), which is a straight line.
2. **Find Points of Intersection:**
- Set the equations equal to each other to find the points where the curves intersect:
\[
8x - x^2 = 2x - 7
\]
- Rearrange the equation:
\[
x^2 - 6x - 7 = 0
\]
- Solve the quadratic equation for \( x \):
\[
x = -1, \quad x = 7
\]
3. **Determine the Bounded Region:**
- The bounded region lies between the points \( x = -1 \) and \( x = 7 \).
4. **Set Up the Integral to Find the Area:**
- The area \( A \) can be found by integrating the difference between the two curves from \( x = -1 \) to \( x = 7 \):
\[
A = \int_{-1}^{7} [(8x - x^2) - (2x - 7)] \, dx
\]
5. **Simplify the Integrand:**
- Simplify the expression inside the integral:
\[
(8x - x^2) - (2x - 7) = 8x - x^2 - 2x + 7 = -x^2 + 6x + 7
\]
6. **Integrate:**
- Evaluate the integral:
\[
A = \int_{-1}^{7} (-x^2 + 6x + 7) \, dx
\]
- Find the antiderivative:
\[
A = \left[ -\frac{x^3}{3} + 3x^2 + 7x \right]_{-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4dca08d-d7c1-4511-81d7-e3b17171b463%2Fdb1a1742-142f-42bd-8453-07aebbe3477d%2Fo4nycml_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 3: Finding Area Bounded by Curves
#### Instructions:
**Graph and then find the area of the region bounded by the curves \( y = 8x - x^2 \) and \( y = 2x - 7 \).**
#### Steps:
1. **Graph the Curves:**
- The first curve is \( y = 8x - x^2 \), which is a downward-opening parabola.
- The second curve is \( y = 2x - 7 \), which is a straight line.
2. **Find Points of Intersection:**
- Set the equations equal to each other to find the points where the curves intersect:
\[
8x - x^2 = 2x - 7
\]
- Rearrange the equation:
\[
x^2 - 6x - 7 = 0
\]
- Solve the quadratic equation for \( x \):
\[
x = -1, \quad x = 7
\]
3. **Determine the Bounded Region:**
- The bounded region lies between the points \( x = -1 \) and \( x = 7 \).
4. **Set Up the Integral to Find the Area:**
- The area \( A \) can be found by integrating the difference between the two curves from \( x = -1 \) to \( x = 7 \):
\[
A = \int_{-1}^{7} [(8x - x^2) - (2x - 7)] \, dx
\]
5. **Simplify the Integrand:**
- Simplify the expression inside the integral:
\[
(8x - x^2) - (2x - 7) = 8x - x^2 - 2x + 7 = -x^2 + 6x + 7
\]
6. **Integrate:**
- Evaluate the integral:
\[
A = \int_{-1}^{7} (-x^2 + 6x + 7) \, dx
\]
- Find the antiderivative:
\[
A = \left[ -\frac{x^3}{3} + 3x^2 + 7x \right]_{-
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