Chapter 5: Problem 60
Use a graphing utility to estimate the value of \(k(k>0)\) so that the region enclosed by \(y=1 /\left(1+k x^{2}\right), y=0, x=0\), and \(x=2\) has an area of 0.6 square unit.
Short Answer
Expert verified
Using a graphing utility shows \( k \approx 0.68 \).
Step by step solution
01
Understand the Problem
The goal is to find the value of \( k \) such that the area under the curve \( y = \frac{1}{1+kx^2} \) from \( x = 0 \) to \( x = 2 \) is 0.6 square units. The region is also bounded by \( y = 0 \).
02
Set Up the Integral
To find the area, set up the integral of the function from \( x = 0 \) to \( x = 2 \). The integral is \( \int_0^2 \frac{1}{1 + kx^2} \, dx \). This integral should equal 0.6.
03
Integrate the Function
The function \( \frac{1}{1+kx^2} \) integrates to \( \frac{1}{\sqrt{k}} \tan^{-1}(\sqrt{k}x) \). Substitute the limits into the indefinite integral to get \[ \frac{1}{\sqrt{k}} (\tan^{-1}(2\sqrt{k}) - \tan^{-1}(0)). \]
04
Solve for k
Set the expression equal to 0.6: \( \frac{1}{\sqrt{k}} \tan^{-1}(2\sqrt{k}) = 0.6 \). This equation needs to be solved for \( k \) either analytically or using a graphing utility to approximate \( k \).
05
Use a Graphing Utility
Enter the function \( f(k) = \frac{1}{\sqrt{k}} \tan^{-1}(2\sqrt{k}) \) into a graphing utility. Plot it and find the value of \( k > 0 \) such that \( f(k) = 0.6 \).
06
Approximate the Value of k
From the graph, estimate the value of \( k \) where the function intersects 0.6. Ensure this \( k \) value is greater than 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a branch of mathematics focused on the concept of integration. Integration is the process of finding the integral of a function, which can be understood as finding the "accumulated total" or "area" under a curve.
Understanding integral calculus provides a powerful tool for solving a wide range of practical and theoretical problems.
- Integrals are often used to find areas, volumes, and other quantities that arise by the addition of infinitesimal increments.
- The process is closely related to differentiation, and together they form the two fundamental operations in calculus.
Understanding integral calculus provides a powerful tool for solving a wide range of practical and theoretical problems.
Definite Integration
Definite integration finds the exact area under a curve between two specific limits. This mathematical operation gives a numerical value representing the area between the curve and the x-axis, within the limits of integration.
- The definite integral is expressed as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated.
- In practice, the definite integral simplifies to a result by evaluating the antiderivative of the function at the upper and lower limits and taking the difference.
Area Under a Curve
Finding the area under a curve is a significant application of integration in calculus, often used to determine the total quantity represented by the curve over a given interval. This calculation is crucial in various fields, including physics, engineering, and economics, where it translates the graphical representation of data into quantitative information.
- The area is determined by integrating the function over the specified range of the x-axis.
- It provides a way to measure how much one quantity changes in relation to another over a particular span.