Chapter 13: Problem 12
Find the function values. \(g(x, y)=\int_{x}^{y} \frac{1}{t} d t\) (a) \(g(4,1)\) (b) \(g(6,3)\)
Short Answer
Expert verified
(a) \(g(4,1) = -\ln4\), (b) \(g(6,3) = -\ln2\)
Step by step solution
01
Understanding the integral
First, we need to compute the integral of the function inside the integral which is \(1/t\). The integral of \(1/t\) with respect to \(t\) is \(\ln|t|\). So, \(g(x, y)\) transforms into \(g(x, y)= \ln|y| - \ln|x|\). When calculating \(g(x, y)\), replace \(x\) and \(y\) with their respective values for each case.
02
Calculate the function value for (a) \(g(4, 1)\)
Substituting \(x=4\) and \(y=1\) into the function \(g(x, y)= \ln|y| - \ln|x|\) we get:\(g(4, 1)= \ln|1| - \ln|4| = \ln1 - \ln4 = -\ln4\).
03
Calculate the function value for (b) \(g(6, 3)\)
Substituting \(x=6\) and \(y=3\) into the functions: \(g(6, 3)= \ln|3| - \ln|6| = \ln3 - \ln6 = -\ln2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm
The natural logarithm, often denoted as \( \ln \), is a logarithm with the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm of a positive number \( t \) symbolizes the power to which \( e \) must be raised to equal \( t \). Mathematically, this is expressed as: - \( e^{\ln(t)} = t \). Understanding the natural logarithm is pivotal in calculus, especially when dealing with integrals of functions involving \( \frac{1}{t} \). The integral of \( \frac{1}{t} \) is \( \ln|t| \). This fact is essential when computing definite integrals that lead to expressions such as \( \ln y - \ln x \).
When evaluating expressions like \( \ln|y| - \ln|x| \), remember that it can also be rewritten as a single logarithm: - \( \ln\left(\frac{|y|}{|x|}\right) \). This property aids in simplifying the computation of definite integrals that result from functions like \( \frac{1}{t} \).
When evaluating expressions like \( \ln|y| - \ln|x| \), remember that it can also be rewritten as a single logarithm: - \( \ln\left(\frac{|y|}{|x|}\right) \). This property aids in simplifying the computation of definite integrals that result from functions like \( \frac{1}{t} \).
Function Evaluation
Function evaluation refers to the process of determining the value of a function at particular inputs, typically numbers. In the context of calculus, this often involves evaluating integrals. For example, to evaluate \( g(x, y) = \int_{x}^{y} \frac{1}{t} dt \), we first must understand what transformations take place from the integral to a computable result.
The integral \( \int \frac{1}{t} dt \) yields \( \ln|t| \). Therefore, the evaluation of the function \( g(x, y) \) becomes \( \ln|y| - \ln|x| \). This evaluation directly involves applying the limits \( x \) and \( y \) into the expression.
The integral \( \int \frac{1}{t} dt \) yields \( \ln|t| \). Therefore, the evaluation of the function \( g(x, y) \) becomes \( \ln|y| - \ln|x| \). This evaluation directly involves applying the limits \( x \) and \( y \) into the expression.
- For \( g(4,1) \), substitute \( x = 4 \) and \( y = 1 \), leading to \( \ln|1| - \ln|4| = -\ln4 \).
- For \( g(6,3) \), substitute \( x = 6 \) and \( y = 3 \), leading to \( \ln|3| - \ln|6| = -\ln2 \).
Calculus Problem Solving
Solving calculus problems often involves a strategic approach. Let's break down the process using the integral \( \int_{x}^{y} \frac{1}{t} dt \) as an example. The primary goal is to evaluate the integral over a defined interval, in this case, from \( x \) to \( y \).
Here are the general steps involved:
By following the structured approach, even more complex integrals and functions can be tackled effectively.
Here are the general steps involved:
- Recognize the function to integrate: Identify that the function \( \frac{1}{t} \) integrates to \( \ln|t| \).
- Understand and apply limits of integration: Substitute the upper and lower limits into the result of the integral. This results in evaluation at \( t = y \) and \( t = x \).
- Simplify: Utilize properties such as \( \ln|y| - \ln|x| = \ln\frac{|y|}{|x|} \) for simplification.
By following the structured approach, even more complex integrals and functions can be tackled effectively.