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:**
*Find the area of the region enclosed by the given curves.*
\[ y = 2 \sqrt{x}, \ y = x - 3, \ x = 0, \ x = 4 \]
To solve this problem, you can follow these steps:
1. **Understand the Curves**:
- The curve \( y = 2 \sqrt{x} \) is a portion of a parabola opening upwards.
- The curve \( y = x - 3 \) is a straight line with a slope of 1, intersecting the y-axis at -3.
- The lines \( x = 0 \) and \( x = 4 \) are vertical lines representing the boundaries of the region.
2. **Find the Points of Intersection**:
- To find where \( y = 2 \sqrt{x} \) and \( y = x - 3 \) intersect, set them equal:
\[ 2 \sqrt{x} = x - 3 \]
Solve this equation to find the x-coordinates of the points of intersection.
3. **Set Up the Integral**:
- The area between two curves from \( x = a \) to \( x = b \) is:
\[ \text{Area} = \int_a^b ( f(x) - g(x) ) \ dx \]
where \( f(x) \) is the upper curve and \( g(x) \) is the lower curve.
- Determine which curve is on top and which is on the bottom within the region \( x = 0 \) to \( x = 4 \).
4. **Calculate the Area**:
- Evaluate the definite integral:
\[ \text{Area} = \int_0^4 ((2 \sqrt{x}) - (x - 3)) \ dx \]
By following these steps, you can find the area of the region enclosed by the given curves.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa216c9ee-ecac-40ec-b583-a361985d4bbd%2F02c43ebf-0758-49f4-917e-95c6866d65a9%2Fu4i9kc4w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
*Find the area of the region enclosed by the given curves.*
\[ y = 2 \sqrt{x}, \ y = x - 3, \ x = 0, \ x = 4 \]
To solve this problem, you can follow these steps:
1. **Understand the Curves**:
- The curve \( y = 2 \sqrt{x} \) is a portion of a parabola opening upwards.
- The curve \( y = x - 3 \) is a straight line with a slope of 1, intersecting the y-axis at -3.
- The lines \( x = 0 \) and \( x = 4 \) are vertical lines representing the boundaries of the region.
2. **Find the Points of Intersection**:
- To find where \( y = 2 \sqrt{x} \) and \( y = x - 3 \) intersect, set them equal:
\[ 2 \sqrt{x} = x - 3 \]
Solve this equation to find the x-coordinates of the points of intersection.
3. **Set Up the Integral**:
- The area between two curves from \( x = a \) to \( x = b \) is:
\[ \text{Area} = \int_a^b ( f(x) - g(x) ) \ dx \]
where \( f(x) \) is the upper curve and \( g(x) \) is the lower curve.
- Determine which curve is on top and which is on the bottom within the region \( x = 0 \) to \( x = 4 \).
4. **Calculate the Area**:
- Evaluate the definite integral:
\[ \text{Area} = \int_0^4 ((2 \sqrt{x}) - (x - 3)) \ dx \]
By following these steps, you can find the area of the region enclosed by the given curves.
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