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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![### Calculating the Area Under the Curve
**Problem Statement:**
Find the area between the graph of \( x = \frac{1}{y^2} \) and the y-axis on the interval [1, 4].
**Diagram:**
The accompanying graph illustrates the function \( x = \frac{1}{y^2} \) within the specified interval. The x-axis spans from -0.5 to 1.5, while the y-axis extends from 0 to 4. The area of interest, which lies between \( y = 1 \) and \( y = 4 \), is shaded in green.
**Explanation:**
To determine the area under the curve \( x = \frac{1}{y^2} \) from \( y = 1 \) to \( y = 4 \), we need to set up and evaluate an integral. Given that the function is defined implicitly, we reframe the integral in terms of \( y \).
The integral to compute the area A is:
\[ A = \int_{1}^{4} \frac{1}{y^2} \, dy \]
**Step-by-Step Solution:**
1. **Set up the Integral:**
To find the area, set up the definite integral of \( x = \frac{1}{y^2} \) from \( y = 1 \) to \( y = 4 \):
\[
A = \int_{1}^{4} \frac{1}{y^2} \, dy
\]
2. **Compute the Integral:**
Evaluate the integral using the antiderivative of \( \frac{1}{y^2} \):
\[
\int \frac{1}{y^2} \, dy = \int y^{-2} \, dy = -y^{-1} = -\frac{1}{y}
\]
3. **Apply the Limits:**
Apply the limits of integration from 1 to 4:
\[
A = \left[ -\frac{1}{y} \right]_{1}^{4} = -\frac{1}{4} - \left( -\frac{1}{1} \right) = -\frac{1}{4} + 1 = \frac{3}{4}
\]
**Conclusion:**
The area between the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bc5b610-4c42-40dc-b81a-6ff9e1fb2857%2F507bf9fb-cb8c-4acf-927d-217b44a67918%2Fw41f9of_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Area Under the Curve
**Problem Statement:**
Find the area between the graph of \( x = \frac{1}{y^2} \) and the y-axis on the interval [1, 4].
**Diagram:**
The accompanying graph illustrates the function \( x = \frac{1}{y^2} \) within the specified interval. The x-axis spans from -0.5 to 1.5, while the y-axis extends from 0 to 4. The area of interest, which lies between \( y = 1 \) and \( y = 4 \), is shaded in green.
**Explanation:**
To determine the area under the curve \( x = \frac{1}{y^2} \) from \( y = 1 \) to \( y = 4 \), we need to set up and evaluate an integral. Given that the function is defined implicitly, we reframe the integral in terms of \( y \).
The integral to compute the area A is:
\[ A = \int_{1}^{4} \frac{1}{y^2} \, dy \]
**Step-by-Step Solution:**
1. **Set up the Integral:**
To find the area, set up the definite integral of \( x = \frac{1}{y^2} \) from \( y = 1 \) to \( y = 4 \):
\[
A = \int_{1}^{4} \frac{1}{y^2} \, dy
\]
2. **Compute the Integral:**
Evaluate the integral using the antiderivative of \( \frac{1}{y^2} \):
\[
\int \frac{1}{y^2} \, dy = \int y^{-2} \, dy = -y^{-1} = -\frac{1}{y}
\]
3. **Apply the Limits:**
Apply the limits of integration from 1 to 4:
\[
A = \left[ -\frac{1}{y} \right]_{1}^{4} = -\frac{1}{4} - \left( -\frac{1}{1} \right) = -\frac{1}{4} + 1 = \frac{3}{4}
\]
**Conclusion:**
The area between the
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