Problem 57 - Essential Calculus

Question

Consider the function \(F(x)=\frac{1}{2} \int_{x}^{x+2} \frac{2}{t^{2}+1} d t\) (a) Write a short paragraph giving a geometric interpretation of the function \(F(x)\) relative to the function \(f(x)=\frac{2}{x^{2}+1}\) Use what you have written to guess the value of \(x\) that will make \(F\) maximum. (b) Perform the specified integration to find an alternative form of \(F(x)\). Use calculus to locate the value of \(x\) that will make \(F\) maximum and compare the result with your guess in part (a).

Step-by-Step Solution

Answer

Based on the solution steps, the geometric interpretation of \(F(x)\) is that it represents half the signed area under the curve \(f(t)\), inside the interval from \(x\) to \(x+2\). The maximum of \(F(x)\) can be guessed to be at \(x=-1\), which is confirmed through calculus in part (b) to be indeed true. The maximum of \(F(x)\) occurs at \(x=-1\).

Step 1: Geometric interpretation and guess

Geometrically, the integral of \(f(t)=\frac{2}{t^{2}+1}\) from \(t=x\) to \(t=x+2\) represents the signed area under the curve \(f(t)\) from \(x\) to \(x+2\). This area is half of the value of \(F(x)\). As you slide the interval of integration along the x-axis, the value of \(F(x)\) changes based on the variations of \(f(t)\). Given the shape of \(f(t)\), which has only one maximum at \(t=0\), it would be expected that \(F(x)\) is maximized when the interval of integration covers as much area under the 'bump' of \(f(t)\). This means \(F(x)\) is most likely maximum around \(x=-1\), where the interval ranges from \(t=-1\) to \(t=1\) covering most of the 'bump'.

Step 2: Evaluation of the Integral

Performing the integration of \(f(t)\) gives the expression \(F(x)=\int_{x}^{x+2} \frac{2}{t^{2}+1} d t\). Upon integration, using the known anti-derivative of the form \(\int \frac{2}{t^{2}+1} dt\), which is \(\arctan(t)\), it yields \(F(x)=\arctan(x+2)-\arctan(x)\)

Step 3: Finding the Maximum of \(F(x)\)

To find where \(F(x)\) reaches its maximum, we need to find the derivative of \(F\), set it equal to zero and solve for \(x\). \(F'(x)=\frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}\). Now, solve the equation \(F'(x)=0\). After some simplification, you find the solutions to be \(x=-1,1\). As \(F(x)\) is a continuous function, we can use the endpoints and critical numbers to apply the First Derivative Test. Evalute \(F(x)\) at these points to conclude that \(F(x)\) reaches its maximum at \(x=-1\), which is consistent with our initial guess.

Key Concepts

Definite Integrals

Definite integrals are fundamental in calculating the area under a curve on a graph between two points along the x-axis. By using the integral of a function, you can determine this area with precision. In the context of

• The exercise given, we analyze the function \(f(t)=\frac{2}{t^{2}+1}\).
• The definite integral \( \int_{x}^{x+2} \frac{2}{t^{2}+1} dt \) represents the area between \(x\) and \(x+2\) under the curve \(f(t)\).

Looking at this definite integral, we're essentially sliding an interval along the graph to measure how the area changes.
The value of \(F(x)\) is also half of the signed area, as mentioned in the original exercise. So, the insight here is that by understanding and computing definite integrals, you can grasp complex relationships between different segments of a curve, such as how they translate into real-world applications like distance, area, or even in calculating some physical quantities.

Maximizing Area

Maximizing area with calculus problems often involves finding the precise interval that covers the most significant section of a graph. The goal, of course, is to capture much of the feature within the bounds of integration.

• In the problem, we recognize that the curve \(f(t)=\frac{2}{t^{2}+1}\) has a prominent rise or 'bump' centered around \(t=0\).
• This means that the integral should ideally start and end at points covering this bump closely.

When you adjust your integration range to cover this feature (which is centered around \(t=0\)), you optimize the area to its maximum possible value under the curve for that given section.
In terms of this problem, predictions like starting the interval at \(x=-1\), so that it spans from \(-1\) to \(1\), ensure that most of the significant area around the bump is captured, hence maximizing \(F(x)\) in the context of the provided function.

Calculus Optimization

Calculus optimization is a technique used to find the maximum or minimum values of a function. The process involves derivatives, as they help determine where a function's slope is zero, indicating potential extremum points.

• First, take the derivative of \(F(x)\), which is \(F'(x)=\frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}\).
• The next step is to set the derivative equal to zero: \(F'(x)=0\), and solve for \(x\).

By simplifying and solving this derivative equation, solutions \(x=-1\) and \(x=1\) are found.
To identify which of these points provides a maximum value for \(F(x)\), the First Derivative Test is applied. This involves plugging these points back into the original function and comparing the resulting values or using the characteristics of continuous functions to predict behavior.
As deduced in both original predictions and confirmed via calculation, \(F(x)\)'s maximum occurs at \(x=-1\), showcasing the accurate use of calculus optimization in real-world function analysis challenges.