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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![### Problem Statement:
**Question 7:**
Find the area of the region bounded by the graphs of \( f(x) = x^2 + 2x + 1 \) and \( g(x) = 3x + 3 \). Write the solution in exact form.
---
### Explanation on How to Approach:
To solve for the area between the curves, you can follow these steps:
1. **Find the Points of Intersection:**
- Set the functions equal to each other to determine the bounds for integration.
\[
x^2 + 2x + 1 = 3x + 3
\]
- Rearrange the equation:
\[
x^2 + 2x + 1 - 3x - 3 = 0 \implies x^2 - x - 2 = 0
\]
- Solve for \(x\):
\[
(x-2)(x+1) = 0
\]
Thus, \( x = 2 \) and \( x = -1 \).
2. **Set Up the Integral to Calculate the Area:**
- The area \( A \) between the two curves from \( x = -1 \) to \( x = 2 \) can be found using the integral:
\[
A = \int_{-1}^{2} (g(x) - f(x)) \, dx
\]
- Substitute the functions \( g(x) \) and \( f(x) \):
\[
A = \int_{-1}^{2} [(3x + 3) - (x^2 + 2x + 1)] \, dx
\]
- Simplify the integrand:
\[
A = \int_{-1}^{2} (3x + 3 - x^2 - 2x - 1) \, dx
\]
\[
A = \int_{-1}^{2} (-x^2 + x + 2) \, dx
\]
3. **Evaluate the Integral:**
- Perform the integration:
\[
\int (-x^2 + x + 2) \, dx = -\frac{x^3}{3} + \frac{x^2}{2} + 2x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e9d947d-304f-415e-95bf-140f5d3b17e4%2F74202fac-e49f-4a0a-825c-600d1d5873fc%2F00phz4j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Question 7:**
Find the area of the region bounded by the graphs of \( f(x) = x^2 + 2x + 1 \) and \( g(x) = 3x + 3 \). Write the solution in exact form.
---
### Explanation on How to Approach:
To solve for the area between the curves, you can follow these steps:
1. **Find the Points of Intersection:**
- Set the functions equal to each other to determine the bounds for integration.
\[
x^2 + 2x + 1 = 3x + 3
\]
- Rearrange the equation:
\[
x^2 + 2x + 1 - 3x - 3 = 0 \implies x^2 - x - 2 = 0
\]
- Solve for \(x\):
\[
(x-2)(x+1) = 0
\]
Thus, \( x = 2 \) and \( x = -1 \).
2. **Set Up the Integral to Calculate the Area:**
- The area \( A \) between the two curves from \( x = -1 \) to \( x = 2 \) can be found using the integral:
\[
A = \int_{-1}^{2} (g(x) - f(x)) \, dx
\]
- Substitute the functions \( g(x) \) and \( f(x) \):
\[
A = \int_{-1}^{2} [(3x + 3) - (x^2 + 2x + 1)] \, dx
\]
- Simplify the integrand:
\[
A = \int_{-1}^{2} (3x + 3 - x^2 - 2x - 1) \, dx
\]
\[
A = \int_{-1}^{2} (-x^2 + x + 2) \, dx
\]
3. **Evaluate the Integral:**
- Perform the integration:
\[
\int (-x^2 + x + 2) \, dx = -\frac{x^3}{3} + \frac{x^2}{2} + 2x
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