Intermediate Algebra
10th Edition
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter9: Functions
Section9.CR: Review Problem Set
Problem 24CR: An outpatient operating room charges each patient a fixed amount per surgery plus an amount per...
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![### Finding the Area of the Shaded Region
Problem Statement:
- Find the area of the shaded region.
Graph Description:
- Two functions are graphed on the coordinate plane:
- The quadratic function \( y = 7x - x^2 \) (in red).
- The linear function \( y = 2x \) (in blue).
- The shaded region is the area between these two curves.
- The curves intersect at the points where \( x \approx 0.534 \) and \( x = 5 \). The point (5, 10) shows one intersection.
### Steps to Solve:
1. **Determine Points of Intersection**:
- Set the equations equal to each other to find where they intersect:
\[
7x - x^2 = 2x
\]
- Rearrange the equation:
\[
x^2 - 5x = 0
\]
- Factor the equation:
\[
x(x - 5) = 0
\]
So, \( x = 0 \) and \( x = 5 \).
2. **Setup the Integral**:
- To find the area between the curves from \( x = 0 \) to \( x = 5 \), use the following integral:
\[
\int_{0}^{5} [(7x - x^2) - 2x] \, dx
\]
- Simplify the integrand:
\[
\int_{0}^{5} (5x - x^2) \, dx
\]
3. **Compute the Integral**:
- Evaluate the integral:
\[
\int (5x - x^2) \, dx = \frac{5x^2}{2} - \frac{x^3}{3}
\]
- Apply the limits from 0 to 5:
\[
\left[\frac{5(5)^2}{2} - \frac{(5)^3}{3}\right] - \left[\frac{5(0)^2}{2} - \frac{(0)^3}{3}\right]
\]
4. **Simplify and Calculate the Area**:
- First part:
\[
\frac{5 \cdot 25}{2}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6653f5ea-9ae4-4965-89e5-4546e31fc625%2Fbfaac841-7855-436f-827a-5a7d7982c905%2Fyqllyd.png&w=3840&q=75)
Transcribed Image Text:### Finding the Area of the Shaded Region
Problem Statement:
- Find the area of the shaded region.
Graph Description:
- Two functions are graphed on the coordinate plane:
- The quadratic function \( y = 7x - x^2 \) (in red).
- The linear function \( y = 2x \) (in blue).
- The shaded region is the area between these two curves.
- The curves intersect at the points where \( x \approx 0.534 \) and \( x = 5 \). The point (5, 10) shows one intersection.
### Steps to Solve:
1. **Determine Points of Intersection**:
- Set the equations equal to each other to find where they intersect:
\[
7x - x^2 = 2x
\]
- Rearrange the equation:
\[
x^2 - 5x = 0
\]
- Factor the equation:
\[
x(x - 5) = 0
\]
So, \( x = 0 \) and \( x = 5 \).
2. **Setup the Integral**:
- To find the area between the curves from \( x = 0 \) to \( x = 5 \), use the following integral:
\[
\int_{0}^{5} [(7x - x^2) - 2x] \, dx
\]
- Simplify the integrand:
\[
\int_{0}^{5} (5x - x^2) \, dx
\]
3. **Compute the Integral**:
- Evaluate the integral:
\[
\int (5x - x^2) \, dx = \frac{5x^2}{2} - \frac{x^3}{3}
\]
- Apply the limits from 0 to 5:
\[
\left[\frac{5(5)^2}{2} - \frac{(5)^3}{3}\right] - \left[\frac{5(0)^2}{2} - \frac{(0)^3}{3}\right]
\]
4. **Simplify and Calculate the Area**:
- First part:
\[
\frac{5 \cdot 25}{2}
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