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...
Related questions
Question
![**Problem Statement:**
Find the area of the region in the first quadrant bounded above by \( y = \sqrt{x} + 2 \) and below by \( y = x \).
**Detailed Explanation of Graph:**
The graph is a two-dimensional plot with \( x \)-axis and \( y \)-axis intersecting at the origin (0,0). The \( x \)-axis ranges from -5 to 5, and the \( y \)-axis ranges from -5 to 5.
The graph includes two curves:
1. The curve \( y = \sqrt{x} + 2 \) is shown in green.
2. The line \( y = x \) is also shown in green.
**Steps to Solve:**
1. **Identify Intersection Points:**
- Set the equations equal to find the points of intersection.
\[
\sqrt{x} + 2 = x
\]
Solve for \( x \):
\[
\sqrt{x} = x - 2
\]
Square both sides:
\[
x = (x - 2)^2 \\
x = x^2 - 4x + 4
\]
Rearrange to form a quadratic equation:
\[
x^2 - 5x + 4 = 0
\]
Factor the quadratic:
\[
(x - 4)(x - 1) = 0
\]
Therefore, \( x = 1 \) and \( x = 4 \).
2. **Calculate the Area:**
- The area is found by integrating the difference of the functions \( f(x) = \sqrt{x} + 2 \) and \( g(x) = x \) between the intersection points \( x = 1 \) and \( x = 4 \).
\[
\text{Area} = \int_{1}^{4} [ (\sqrt{x} + 2) - x ] \, dx
\]
Simplify the integrand:
\[
\text{Area} = \int_{1}^{4} (\sqrt{x} + 2 - x) \, dx
\]
Compute the integral:
\[
= \int_{1}^{4} \sqrt{x} \, dx + \int_{1}^{4}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bc5b610-4c42-40dc-b81a-6ff9e1fb2857%2Fa06b2023-4861-4034-adc4-41bdf5abf285%2Ffw1ic1_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area of the region in the first quadrant bounded above by \( y = \sqrt{x} + 2 \) and below by \( y = x \).
**Detailed Explanation of Graph:**
The graph is a two-dimensional plot with \( x \)-axis and \( y \)-axis intersecting at the origin (0,0). The \( x \)-axis ranges from -5 to 5, and the \( y \)-axis ranges from -5 to 5.
The graph includes two curves:
1. The curve \( y = \sqrt{x} + 2 \) is shown in green.
2. The line \( y = x \) is also shown in green.
**Steps to Solve:**
1. **Identify Intersection Points:**
- Set the equations equal to find the points of intersection.
\[
\sqrt{x} + 2 = x
\]
Solve for \( x \):
\[
\sqrt{x} = x - 2
\]
Square both sides:
\[
x = (x - 2)^2 \\
x = x^2 - 4x + 4
\]
Rearrange to form a quadratic equation:
\[
x^2 - 5x + 4 = 0
\]
Factor the quadratic:
\[
(x - 4)(x - 1) = 0
\]
Therefore, \( x = 1 \) and \( x = 4 \).
2. **Calculate the Area:**
- The area is found by integrating the difference of the functions \( f(x) = \sqrt{x} + 2 \) and \( g(x) = x \) between the intersection points \( x = 1 \) and \( x = 4 \).
\[
\text{Area} = \int_{1}^{4} [ (\sqrt{x} + 2) - x ] \, dx
\]
Simplify the integrand:
\[
\text{Area} = \int_{1}^{4} (\sqrt{x} + 2 - x) \, dx
\]
Compute the integral:
\[
= \int_{1}^{4} \sqrt{x} \, dx + \int_{1}^{4}
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