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Find the area between the graph of x = and the y-axis on the interval [1, 4]. 4 3 2- 1 0.5 05 1 1.5 4

Find the area between the graph of x = and the y-axis on the interval [1, 4]. 4 3 2- 1 0.5 05 1 1.5 4

Calculus: Early Transcendentals
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
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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