Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter9: Polynomial And Rational Functions
Section9.4: Graphing Polynomial Functions
Problem 43PS
Related questions
Question
![**Problem Statement:**
Find the area of the region enclosed by the curves \( y = x^3 \) and \( y = 4x \).
**Solution:**
To solve this problem, follow these steps:
1. **Find the Points of Intersection:**
- Set the equations equal to each other to find the points where the curves intersect:
\[
x^3 = 4x
\]
- Solve for \( x \):
\[
x^3 - 4x = 0
\]
\[
x(x^2 - 4) = 0
\]
\[
x(x - 2)(x + 2) = 0
\]
So, the solutions are \( x = 0 \), \( x = 2 \), and \( x = -2 \).
2. **Set Up the Integral:**
- The area between the curves is found by integrating the difference between the functions over the interval from \( x = -2 \) to \( x = 2 \).
- The upper curve is \( y = 4x \) and the lower curve is \( y = x^3 \).
3. **Calculate the Integral:**
- The area can be calculated as:
\[
\text{Area} = \int_{-2}^{2} (4x - x^3) \, dx
\]
- Integrate term by term:
\[
\int_{-2}^{2} 4x \, dx - \int_{-2}^{2} x^3 \, dx
\]
4. **Evaluate the Integrals:**
- For the first integral:
\[
\int 4x \, dx = 2x^2
\]
Evaluate from \(-2\) to \(2\):
\[
2(2)^2 - 2(-2)^2 = 2(4) - 2(4) = 8 - 8 = 0
\]
- For the second integral:
\[
\int x^3 \, dx = \frac{x^4}{4}
\]
Evaluate from \(-2\) to \(2\):
\[
\left[ \frac{(2)^4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde326383-8b00-4076-b3f6-1f860c5e4029%2F08101e22-5d3a-4bcb-a06f-2f16dce027dc%2Fbmgh7i_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area of the region enclosed by the curves \( y = x^3 \) and \( y = 4x \).
**Solution:**
To solve this problem, follow these steps:
1. **Find the Points of Intersection:**
- Set the equations equal to each other to find the points where the curves intersect:
\[
x^3 = 4x
\]
- Solve for \( x \):
\[
x^3 - 4x = 0
\]
\[
x(x^2 - 4) = 0
\]
\[
x(x - 2)(x + 2) = 0
\]
So, the solutions are \( x = 0 \), \( x = 2 \), and \( x = -2 \).
2. **Set Up the Integral:**
- The area between the curves is found by integrating the difference between the functions over the interval from \( x = -2 \) to \( x = 2 \).
- The upper curve is \( y = 4x \) and the lower curve is \( y = x^3 \).
3. **Calculate the Integral:**
- The area can be calculated as:
\[
\text{Area} = \int_{-2}^{2} (4x - x^3) \, dx
\]
- Integrate term by term:
\[
\int_{-2}^{2} 4x \, dx - \int_{-2}^{2} x^3 \, dx
\]
4. **Evaluate the Integrals:**
- For the first integral:
\[
\int 4x \, dx = 2x^2
\]
Evaluate from \(-2\) to \(2\):
\[
2(2)^2 - 2(-2)^2 = 2(4) - 2(4) = 8 - 8 = 0
\]
- For the second integral:
\[
\int x^3 \, dx = \frac{x^4}{4}
\]
Evaluate from \(-2\) to \(2\):
\[
\left[ \frac{(2)^4
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