Chapter 6: Problem 33
Find an antiderivative \(F(x)\) with \(F^{\prime}(x)=\) \(f(x)\) and \(F(0)=0 .\) Is there only one possible solution? $$f(x)=\frac{1}{4} x$$
Short Answer
Expert verified
The antiderivative is \(F(x) = \frac{x^2}{8}\), and it is the only solution satisfying the conditions.
Step by step solution
01
Understand the Problem
We need to find an antiderivative \(F(x)\) such that \(F'(x) = \frac{1}{4}x\) and also satisfies the condition \(F(0) = 0\). We are asked to determine if there's only one possible solution for \(F(x)\).
02
Find the General Antiderivative
To find the general antiderivative, we need to integrate \(f(x) = \frac{1}{4}x\). The antiderivative of \(x\) is \(\frac{x^2}{2}\), so the antiderivative of \(\frac{1}{4}x\) is \(\frac{1}{4} \cdot \frac{x^2}{2} = \frac{x^2}{8}\). Therefore, the general antiderivative is \(F(x) = \frac{x^2}{8} + C\), where \(C\) is an arbitrary constant.
03
Apply the Initial Condition
We need \(F(0) = 0\). Substitute \(x = 0\) into the general antiderivative: \(F(0) = \frac{0^2}{8} + C = C\). Since \(F(0) = 0\), it follows that \(C = 0\).
04
Determine the Specific Solution
Using the value of \(C\) found in the previous step, we have \(F(x) = \frac{x^2}{8}\). This is a specific function that satisfies both \(F'(x) = \frac{1}{4}x\) and \(F(0) = 0\).
05
Analyze Uniqueness
In general, an antiderivative can have a family of functions differing by a constant, but the condition \(F(0) = 0\) uniquely determines that \(C = 0\). Therefore, this is the only function that satisfies all conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
initial conditions
When we work with antiderivatives, initial conditions play a crucial role in finding a particular solution from a family of solutions. An initial condition is a specific requirement, often given by a value that the solution must satisfy at a certain point.
This specific value helps us determine the constant of integration, which is typically a constant added to the indefinite integral. In our exercise, the initial condition given is \(F(0) = 0\).
This means that the function \(F(x)\), when evaluated at \(x = 0\), must yield 0. Without such an initial condition, we could not uniquely determine the constant \(C\). Instead, we would have a family of solutions, each differing by a constant value.
This specific value helps us determine the constant of integration, which is typically a constant added to the indefinite integral. In our exercise, the initial condition given is \(F(0) = 0\).
This means that the function \(F(x)\), when evaluated at \(x = 0\), must yield 0. Without such an initial condition, we could not uniquely determine the constant \(C\). Instead, we would have a family of solutions, each differing by a constant value.
unique solution
In the context of antiderivatives, a unique solution arises when initial conditions allow us to pin down the exact value of the constant of integration. This is essential because, generally speaking, the indefinite integral offers a broad set of solutions.
These solutions show up in the form \(F(x) = \frac{x^2}{8} + C\), where \(C\) is an arbitrary constant. Without an initial condition, \(C\) can be any real number, leading to multiple possible functions. However, applying the initial condition \(F(0) = 0\) restricts \(C\) to a specific value.
In this case, applying \(F(0) = 0\) implies that \(C\) must be 0, which then results in the unique solution \(F(x) = \frac{x^2}{8}\). Thus, the initial condition ensures that there is not just any solution, but a unique solution that meets the criteria given.
These solutions show up in the form \(F(x) = \frac{x^2}{8} + C\), where \(C\) is an arbitrary constant. Without an initial condition, \(C\) can be any real number, leading to multiple possible functions. However, applying the initial condition \(F(0) = 0\) restricts \(C\) to a specific value.
In this case, applying \(F(0) = 0\) implies that \(C\) must be 0, which then results in the unique solution \(F(x) = \frac{x^2}{8}\). Thus, the initial condition ensures that there is not just any solution, but a unique solution that meets the criteria given.
integration
Integration is a fundamental process in calculus used to find antiderivatives. It essentially involves finding a function whose derivative is the given function. This is the reverse operation of differentiation.
In our exercise, the task is to integrate the function \(f(x) = \frac{1}{4}x\). Integration results in the general form of the antiderivative, \(F(x)\).
To integrate \(f(x)\), we calculate the indefinite integral: \[ \int \frac{1}{4} x \, dx = \frac{1}{4} \cdot \frac{x^2}{2} = \frac{x^2}{8}\].
The integral also includes the constant of integration, \(C\), representing the family of all antiderivatives of a given function. Hence, the general form obtained is \(F(x) = \frac{x^2}{8} + C\). Integration transforms our initial function into this broader solution, which can then be refined using initial conditions to find a unique function when required.
In our exercise, the task is to integrate the function \(f(x) = \frac{1}{4}x\). Integration results in the general form of the antiderivative, \(F(x)\).
To integrate \(f(x)\), we calculate the indefinite integral: \[ \int \frac{1}{4} x \, dx = \frac{1}{4} \cdot \frac{x^2}{2} = \frac{x^2}{8}\].
The integral also includes the constant of integration, \(C\), representing the family of all antiderivatives of a given function. Hence, the general form obtained is \(F(x) = \frac{x^2}{8} + C\). Integration transforms our initial function into this broader solution, which can then be refined using initial conditions to find a unique function when required.