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
![**Problem Statement**
Sketch the region enclosed by the equations \( y = 2x \) and \( y = 9x^2 \). Determine whether to integrate with respect to \( x \) or \( y \). Then, calculate the area of the region.
**Solution Approach**
To solve this problem:
1. **Graph the Curves:**
- Plot the linear function \( y = 2x \), which is a straight line with a slope of 2, passing through the origin.
- Plot the quadratic function \( y = 9x^2 \), which is a parabola opening upwards with its vertex at the origin.
2. **Find the Points of Intersection:**
- Set the equations equal to each other to find the points where the graphs intersect:
\[
2x = 9x^2 \implies 9x^2 - 2x = 0 \implies x(9x - 2) = 0
\]
- Solve for \( x \): \( x = 0 \) or \( x = \frac{2}{9} \).
- Therefore, the points of intersection are \( (0,0) \) and \( \left( \frac{2}{9}, \frac{4}{9} \right) \).
3. **Decide on the Variable of Integration:**
- Integrating with respect to \( x \) is simpler here since the functions are already expressed in terms of \( x \).
4. **Setup the Integral to Find the Area:**
- The area \( A \) can be found by integrating the difference of the functions over the interval \([0, \frac{2}{9}]\):
\[
A = \int_{0}^{\frac{2}{9}} (2x - 9x^2) \, dx
\]
5. **Evaluate the Integral:**
- Compute the definite integral:
\[
A = \left[ x^2 - 3x^3 \right]_{0}^{\frac{2}{9}}
\]
- Calculate the values:
\[
A = \left( \left( \frac{2}{9} \right)^2 - 3 \left( \frac{2}{9} \right)^3 \right) - (0^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F33eba9e5-eb7d-44a1-93e3-ed0ee897e17f%2F57716094-901b-40b6-9c7c-72638cd73604%2Ff1jcl7s_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Sketch the region enclosed by the equations \( y = 2x \) and \( y = 9x^2 \). Determine whether to integrate with respect to \( x \) or \( y \). Then, calculate the area of the region.
**Solution Approach**
To solve this problem:
1. **Graph the Curves:**
- Plot the linear function \( y = 2x \), which is a straight line with a slope of 2, passing through the origin.
- Plot the quadratic function \( y = 9x^2 \), which is a parabola opening upwards with its vertex at the origin.
2. **Find the Points of Intersection:**
- Set the equations equal to each other to find the points where the graphs intersect:
\[
2x = 9x^2 \implies 9x^2 - 2x = 0 \implies x(9x - 2) = 0
\]
- Solve for \( x \): \( x = 0 \) or \( x = \frac{2}{9} \).
- Therefore, the points of intersection are \( (0,0) \) and \( \left( \frac{2}{9}, \frac{4}{9} \right) \).
3. **Decide on the Variable of Integration:**
- Integrating with respect to \( x \) is simpler here since the functions are already expressed in terms of \( x \).
4. **Setup the Integral to Find the Area:**
- The area \( A \) can be found by integrating the difference of the functions over the interval \([0, \frac{2}{9}]\):
\[
A = \int_{0}^{\frac{2}{9}} (2x - 9x^2) \, dx
\]
5. **Evaluate the Integral:**
- Compute the definite integral:
\[
A = \left[ x^2 - 3x^3 \right]_{0}^{\frac{2}{9}}
\]
- Calculate the values:
\[
A = \left( \left( \frac{2}{9} \right)^2 - 3 \left( \frac{2}{9} \right)^3 \right) - (0^2
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