2. Which integral gives the area of the region in the first quadrant bounded by the graphs of y2 = x - 1, x = 0, y = 0, and y = 2? 5 O {2dx+ (2-vx-1)dx 01(2-(x-1)0x О j2ax+ (3-x)dx О |2ax- (2-vx-1)dx
2. Which integral gives the area of the region in the first quadrant bounded by the graphs of y2 = x - 1, x = 0, y = 0, and y = 2? 5 O {2dx+ (2-vx-1)dx 01(2-(x-1)0x О j2ax+ (3-x)dx О |2ax- (2-vx-1)dx
Calculus: Early Transcendentals
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Chapter1: Functions And Models
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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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![### Calculus Problem: Finding the Area Bounded by Curves
**Problem Statement:**
Determine which integral provides the area of the region in the first quadrant bounded by the graphs of \(y^2 = x - 1\), \(x = 0\), \(y = 0\), and \(y = 2\).
**Options:**
1. \[
\int_{0}^{1} 2 \, dx + \int_{1}^{5} \left(2 - \sqrt{x-1} \right) dx
\]
2. \[
\int_{0}^{5} \left(2 - \sqrt{x-1} \right) dx
\]
3. \[
\int_{0}^{1} 2 \, dx + \int_{1}^{5} \left(3 - x \right) dx
\]
4. \[
\int_{0}^{1} 2 \, dx - \int_{1}^{5} \left(2 - \sqrt{x-1} \right) dx
\]
**Explanation:**
Each option presents a different way to calculate the area using integral calculus. To solve this problem, we need to consider the boundaries provided in the problem and set up the appropriate integral(s):
- **\(y^2 = x - 1\)** is the primary curve.
- **\(x = 0\)** and **\(y = 0\)** represent the y-axis and x-axis, respectively.
- **\(y = 2\)** is the horizontal line at \(y = 2\).
When these curves and lines intersect and bound a region, the area can be found by integrating the appropriate functions:
1. **Breaking integral at \(x = 1\)**:
- From \(x = 0\) to \(x = 1\), the constant function \(y = 2\) bounds the area.
- From \(x = 1\) to \(x = 5\), the function \(y = \sqrt{x-1}\) is bounded above by the line \(y = 2\).
Given this understanding, the correct integral that provides the area of the region in the first quadrant bounded by the given graphs is:
\[
\int_{0}^{1} 2 \, dx + \int_{1}^{5} \left(2 - \sqrt{x-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4bf8d9f8-9f3c-4f11-9f6a-4feeeac00eb3%2Fa560f31c-b927-465c-8e58-c2e93756016c%2Ft4lmbof_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Finding the Area Bounded by Curves
**Problem Statement:**
Determine which integral provides the area of the region in the first quadrant bounded by the graphs of \(y^2 = x - 1\), \(x = 0\), \(y = 0\), and \(y = 2\).
**Options:**
1. \[
\int_{0}^{1} 2 \, dx + \int_{1}^{5} \left(2 - \sqrt{x-1} \right) dx
\]
2. \[
\int_{0}^{5} \left(2 - \sqrt{x-1} \right) dx
\]
3. \[
\int_{0}^{1} 2 \, dx + \int_{1}^{5} \left(3 - x \right) dx
\]
4. \[
\int_{0}^{1} 2 \, dx - \int_{1}^{5} \left(2 - \sqrt{x-1} \right) dx
\]
**Explanation:**
Each option presents a different way to calculate the area using integral calculus. To solve this problem, we need to consider the boundaries provided in the problem and set up the appropriate integral(s):
- **\(y^2 = x - 1\)** is the primary curve.
- **\(x = 0\)** and **\(y = 0\)** represent the y-axis and x-axis, respectively.
- **\(y = 2\)** is the horizontal line at \(y = 2\).
When these curves and lines intersect and bound a region, the area can be found by integrating the appropriate functions:
1. **Breaking integral at \(x = 1\)**:
- From \(x = 0\) to \(x = 1\), the constant function \(y = 2\) bounds the area.
- From \(x = 1\) to \(x = 5\), the function \(y = \sqrt{x-1}\) is bounded above by the line \(y = 2\).
Given this understanding, the correct integral that provides the area of the region in the first quadrant bounded by the given graphs is:
\[
\int_{0}^{1} 2 \, dx + \int_{1}^{5} \left(2 - \sqrt{x-
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