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 between the graph of \( x = \frac{1}{y^2} \) and the y-axis on the interval \([1, 4]\).
## Explanation
In this problem, we need to determine the area between the curve given by the equation \( x = \frac{1}{y^2} \) and the y-axis, constrained within the interval from \( y = 1 \) to \( y = 4 \).
### Graph Description
The graph provided shows the curve \( x = \frac{1}{y^2} \), which is a plot in the first quadrant of the coordinate system with the following characteristics:
- The \( x \)-axis ranges approximately from -0.5 to 1.5.
- The \( y \)-axis ranges from 0 to 4.
- The curve starts curving sharply near the y-axis at the lower end of the interval (near \( y = 1 \)) and becomes less steep as y increases (towards \( y = 4 \)).
### Shaded Area
The region of interest for finding the area is the part between this curve and the y-axis on the interval \([1, 4]\). This region is shaded in green on the graph.
### Solution
To determine this area, we need to set up and evaluate the definite integral of the function \( x = \frac{1}{y^2} \) over the given interval:
\[ \text{Area} = \int_{1}^{4} \frac{1}{y^2} \, dy \]
The integral can be solved by:
1. Finding the antiderivative of \( \frac{1}{y^2} \),
2. Evaluating the antiderivative at the bounds of integration \( y = 1 \) and \( y = 4 \),
3. Subtracting the values to get the total area.
### Calculation
1. The antiderivative of \( \frac{1}{y^2} \) is:
\[ \int \frac{1}{y^2} \, dy = -\frac{1}{y} \]
2. Evaluating this antiderivative at the upper and lower limits:
\[ \left[-\frac{1}{y}\right]_{1}^{4} = -\frac{1}{4} - \left(-\frac{1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5cd77d14-523b-4d70-a9c2-c3967239ea67%2F47e53a9c-58d7-45f8-9ac9-4c3035258073%2F977s79_processed.png&w=3840&q=75)
Transcribed Image Text:## Problem Statement
Find the area between the graph of \( x = \frac{1}{y^2} \) and the y-axis on the interval \([1, 4]\).
## Explanation
In this problem, we need to determine the area between the curve given by the equation \( x = \frac{1}{y^2} \) and the y-axis, constrained within the interval from \( y = 1 \) to \( y = 4 \).
### Graph Description
The graph provided shows the curve \( x = \frac{1}{y^2} \), which is a plot in the first quadrant of the coordinate system with the following characteristics:
- The \( x \)-axis ranges approximately from -0.5 to 1.5.
- The \( y \)-axis ranges from 0 to 4.
- The curve starts curving sharply near the y-axis at the lower end of the interval (near \( y = 1 \)) and becomes less steep as y increases (towards \( y = 4 \)).
### Shaded Area
The region of interest for finding the area is the part between this curve and the y-axis on the interval \([1, 4]\). This region is shaded in green on the graph.
### Solution
To determine this area, we need to set up and evaluate the definite integral of the function \( x = \frac{1}{y^2} \) over the given interval:
\[ \text{Area} = \int_{1}^{4} \frac{1}{y^2} \, dy \]
The integral can be solved by:
1. Finding the antiderivative of \( \frac{1}{y^2} \),
2. Evaluating the antiderivative at the bounds of integration \( y = 1 \) and \( y = 4 \),
3. Subtracting the values to get the total area.
### Calculation
1. The antiderivative of \( \frac{1}{y^2} \) is:
\[ \int \frac{1}{y^2} \, dy = -\frac{1}{y} \]
2. Evaluating this antiderivative at the upper and lower limits:
\[ \left[-\frac{1}{y}\right]_{1}^{4} = -\frac{1}{4} - \left(-\frac{1
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