Find the area between x = y(y - 1) and y = shown in the graph below. Show how to set up the integral, but you may use your calculator to evaluate it. x = y(y-1) y=x/2 6 (6.3)
Find the area between x = y(y - 1) and y = shown in the graph below. Show how to set up the integral, but you may use your calculator to evaluate it. x = y(y-1) y=x/2 6 (6.3)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem 8: Area Between Curves
**Problem Statement:**
Find the area between \( x = y(y - 1) \) and \( y = \frac{x}{2} \), shown in the graph below. Show how to set up the integral, but you may use your calculator to evaluate it.
**Solution Steps:**
1. **Identify the curves:**
- The first equation is \( x = y(y - 1) \). This can be rewritten as \( x = y^2 - y \).
- The second equation is \( y = \frac{x}{2} \).
2. **Find the points of intersection:**
- To find the points of intersection, set \( y = \frac{x}{2} \) into \( x = y(y - 1) \).
- Substitute \( y = \frac{x}{2} \) into \( x = y^2 - y \) to get:
\[
x = \left(\frac{x}{2}\right)^2 - \frac{x}{2}
\]
- Simplify and solve for \( x \):
\[
x = \frac{x^2}{4} - \frac{x}{2}
\]
\[
\frac{x^2}{4} - \frac{3x}{2} = 0
\]
\[
x^2 - 6x = 0
\]
\[
x(x - 6) = 0
\]
- This gives \( x = 0 \) and \( x = 6 \). For these \( x \)-values, the corresponding \( y \)-values are \( y = 0 \) and \( y = 3 \). Hence, the points of intersection are \( (0, 0) \) and \( (6, 3) \).
3. **Set up the integral:**
- The area \( A \) between the curves is given by the integral of the top function minus the bottom function over the interval of \( x \)-values from \( 0 \) to \( 6 \).
\[
A = \int_{0}^{6} \left( \frac{x}{2} - \left( \sqrt{x+1} - 1 \right) \right) dx
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff015db4-0da8-42fc-9d60-01a672218c35%2F6fde0d97-9c49-42ad-acab-4a62693bc4c1%2Frek2az9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 8: Area Between Curves
**Problem Statement:**
Find the area between \( x = y(y - 1) \) and \( y = \frac{x}{2} \), shown in the graph below. Show how to set up the integral, but you may use your calculator to evaluate it.
**Solution Steps:**
1. **Identify the curves:**
- The first equation is \( x = y(y - 1) \). This can be rewritten as \( x = y^2 - y \).
- The second equation is \( y = \frac{x}{2} \).
2. **Find the points of intersection:**
- To find the points of intersection, set \( y = \frac{x}{2} \) into \( x = y(y - 1) \).
- Substitute \( y = \frac{x}{2} \) into \( x = y^2 - y \) to get:
\[
x = \left(\frac{x}{2}\right)^2 - \frac{x}{2}
\]
- Simplify and solve for \( x \):
\[
x = \frac{x^2}{4} - \frac{x}{2}
\]
\[
\frac{x^2}{4} - \frac{3x}{2} = 0
\]
\[
x^2 - 6x = 0
\]
\[
x(x - 6) = 0
\]
- This gives \( x = 0 \) and \( x = 6 \). For these \( x \)-values, the corresponding \( y \)-values are \( y = 0 \) and \( y = 3 \). Hence, the points of intersection are \( (0, 0) \) and \( (6, 3) \).
3. **Set up the integral:**
- The area \( A \) between the curves is given by the integral of the top function minus the bottom function over the interval of \( x \)-values from \( 0 \) to \( 6 \).
\[
A = \int_{0}^{6} \left( \frac{x}{2} - \left( \sqrt{x+1} - 1 \right) \right) dx
\]
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