Problem 68
Question
A definite integral is given. Do not attempt to calculate its value \(V .\) Instead, find the extreme values of the integrand on the interval of integration, and use these extreme values together with the inequalities of line (5.3.5) to obtain numbers \(A\) and \(B\) such that \(A \leq V \leq B\). $$ \int_{1 / 2}^{1} e^{-x} \ln (1+x) d x $$
Step-by-Step Solution
Verified Answer
The integral is bounded by \( A = e^{-1/2} \ln\left(\frac{3}{2}\right) \times \frac{1}{2} \) and \( B = e^{-1} \ln(2) \times \frac{1}{2} \).
1Step 1: Understand the Problem
We need to find the extreme values of the function \( f(x) = e^{-x} \ln(1+x) \) on the interval \( \left[ \frac{1}{2}, 1 \right] \), and then use these extreme values to bound the definite integral \( \int_{1/2}^{1} e^{-x} \ln(1+x) \, dx \) between two numbers \( A \) and \( B \), such that \( A \leq V \leq B \).
2Step 2: Find the Derivative of the Integrand
The function \( f(x) = e^{-x} \ln(1+x) \) is the integrand. First, we find its derivative using the product rule. Let \( u(x) = e^{-x} \) and \( v(x) = \ln(1+x) \). Then, \( f'(x) = u'(x)v(x) + u(x)v'(x) \). Calculate each: \( u'(x) = -e^{-x} \) and \( v'(x) = \frac{1}{1+x} \). Therefore, \( f'(x) = -e^{-x} \ln(1+x) + \frac{e^{-x}}{1+x} \).
3Step 3: Solve the Derivative Equal to Zero
Solving \( f'(x) = 0 \) will help find the critical points. Set \(-e^{-x} \ln(1+x) + \frac{e^{-x}}{1+x} = 0\). Simplifying, we get \( \ln(1+x) = \frac{1}{1+x} \). This is a transcendental equation that might require numerical methods for an exact solution; however, estimating or further analysis could be used for approximate solutions.
4Step 4: Evaluate the Function at Critical Points and Endpoints
Evaluate \( f(x) \) at the critical point(s) found in the solution above (if solvable), and at the endpoints \( x = \frac{1}{2} \) and \( x = 1 \). For the endpoints: \( f\left(\frac{1}{2}\right) = e^{-1/2} \ln\left(\frac{3}{2}\right) \) and \( f(1) = e^{-1} \ln(2) \).
5Step 5: Use Extreme Values to Find Bounds
Identify the maximum and minimum values of \( f(x) \) on \( \left[ \frac{1}{2}, 1 \right] \) among the points evaluated. Suppose \( m \) is the minimum and \( M \) is the maximum value. Then by the inequality \( m(b-a) \leq \int_{a}^{b} f(x) \, dx \leq M(b-a) \), we find \( A = m \left( \frac{1}{2} \right) \) and \( B = M \left( \frac{1}{2} \right) \).
6Step 6: Conclude with Bounds
Using the function values found, compute \( A = e^{-1/2} \ln\left(\frac{3}{2}\right) \times \frac{1}{2} \) and \( B = e^{-1} \ln(2) \times \frac{1}{2} \), assuming exact points were determined correctly. These are the bounds for the integral: \( A \leq V \leq B \).
Key Concepts
Extreme ValuesInequalitiesProduct RuleTranscendental Equation
Extreme Values
Extreme values of a function are simply the highest and lowest values that the function can take on a given interval. When dealing with definite integrals, finding these extreme values can help us understand the behavior of the integrand over the interval and set bounds on the integral's value.
To determine these extreme values, we first calculate the derivative of the function, as this will allow us to find critical points. Critical points occur where the derivative is zero or undefined, indicating potential local maxima or minima. In our case, we use the function \( f(x) = e^{-x} \ln(1+x) \) on the interval \( \left[ \frac{1}{2}, 1 \right] \). By evaluating the function at these critical points and at the endpoints of the interval, we identify where the extreme values occur.
By comparing these values, we can determine the minimum and maximum values \( m \) and \( M \) on the interval. These extreme function values provide us with the key elements needed to bound the definite integral.
To determine these extreme values, we first calculate the derivative of the function, as this will allow us to find critical points. Critical points occur where the derivative is zero or undefined, indicating potential local maxima or minima. In our case, we use the function \( f(x) = e^{-x} \ln(1+x) \) on the interval \( \left[ \frac{1}{2}, 1 \right] \). By evaluating the function at these critical points and at the endpoints of the interval, we identify where the extreme values occur.
By comparing these values, we can determine the minimum and maximum values \( m \) and \( M \) on the interval. These extreme function values provide us with the key elements needed to bound the definite integral.
Inequalities
Inequalities are mathematical expressions used to show the relative size or order of two values. In the context of definite integrals, inequalities play a significant role in bounding the integral's value between two numbers. Specifically, for an integrable function \( f(x) \) over an interval \([a, b]\), the inequality \( m(b-a) \leq \int_{a}^{b} f(x) \, dx \leq M(b-a) \) can be applied.
Here, \( m \) and \( M \) represent the minimum and maximum values of the function on the given interval. We multiply these extreme values by the length of the interval \( (b-a) \) to estimate the lower and upper bounds of the integral. This approach provides a practical way to assess the range of possible integral values without performing actual integration.
Using inequalities, we can make educated estimates about the range in which the definite integral lies. This is particularly useful when an exact integral value is hard to compute or unnecessary for the problem at hand.
Here, \( m \) and \( M \) represent the minimum and maximum values of the function on the given interval. We multiply these extreme values by the length of the interval \( (b-a) \) to estimate the lower and upper bounds of the integral. This approach provides a practical way to assess the range of possible integral values without performing actual integration.
Using inequalities, we can make educated estimates about the range in which the definite integral lies. This is particularly useful when an exact integral value is hard to compute or unnecessary for the problem at hand.
Product Rule
The product rule is an essential tool in calculus for finding the derivative of a product of two functions. It states that if \( u(x) \) and \( v(x) \) are two differentiable functions, then the derivative of their product \( f(x) = u(x)v(x) \) can be expressed as \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
Applying the product rule allows us to handle more complex functions that are products of simpler ones. In our problem, the integrand \( f(x) = e^{-x} \ln(1+x) \) is a product of the functions \( e^{-x} \) and \( \ln(1+x) \).
Using the product rule, we find the derivative \( f'(x) \) to locate critical points which are necessary to determine extreme values. It simplifies the differentiation process and provides the necessary setup for further analysis of the function behavior. This is crucial for understanding the complete picture of the function across the defined interval.
Applying the product rule allows us to handle more complex functions that are products of simpler ones. In our problem, the integrand \( f(x) = e^{-x} \ln(1+x) \) is a product of the functions \( e^{-x} \) and \( \ln(1+x) \).
Using the product rule, we find the derivative \( f'(x) \) to locate critical points which are necessary to determine extreme values. It simplifies the differentiation process and provides the necessary setup for further analysis of the function behavior. This is crucial for understanding the complete picture of the function across the defined interval.
Transcendental Equation
A transcendental equation is an equation containing transcendental functions, which include functions like exponential, logarithmic, trigonometric, etc. These equations often do not have closed-form solutions, making them challenging to solve analytically.
In our exercise, after finding the derivative \(-e^{-x} \ln(1+x) + \frac{e^{-x}}{1+x} = 0\), we end up with a transcendental equation \( \ln(1+x) = \frac{1}{1+x} \). Solving such equations typically requires numerical methods or approximations since analytical solutions might be complex or impossible to achieve.
The approach to handling transcendental equations is through iteration or approximations, such as the Newton-Raphson method or other numerical strategies. While dealing with these equations might seem daunting, the focus should be on understanding their structure and using available techniques to explore potential solutions.
In calculus and beyond, getting comfortable with transcendental equations is valuable, as it aligns closely with solving practical, real-life problem scenarios.
In our exercise, after finding the derivative \(-e^{-x} \ln(1+x) + \frac{e^{-x}}{1+x} = 0\), we end up with a transcendental equation \( \ln(1+x) = \frac{1}{1+x} \). Solving such equations typically requires numerical methods or approximations since analytical solutions might be complex or impossible to achieve.
The approach to handling transcendental equations is through iteration or approximations, such as the Newton-Raphson method or other numerical strategies. While dealing with these equations might seem daunting, the focus should be on understanding their structure and using available techniques to explore potential solutions.
In calculus and beyond, getting comfortable with transcendental equations is valuable, as it aligns closely with solving practical, real-life problem scenarios.
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