Problem 27
Question
Find an antiderivative \(F(x)\) with \(F^{\prime}(x)=\) \(f(x)\) and \(F(0)=0\). Is there only one possible solution? $$ f(x)=3 $$
Step-by-Step Solution
Verified Answer
The antiderivative is \(F(x) = 3x\). There is only one solution because of the initial condition \(F(0)=0\).
1Step 1: Recognize the Problem Type
We are tasked with finding an antiderivative, which means we need to determine a function whose derivative is the given function \(f(x)\).
2Step 2: Identify the Given Function
The given function is \(f(x) = 3\).
3Step 3: Find the General Antiderivative
The antiderivative of a constant \(c\) is \(F(x) = cx + C\), where \(C\) is a constant. Therefore, the general antiderivative of \(3\) is \(F(x) = 3x + C\).
4Step 4: Apply the Initial Condition
We are given the initial condition \(F(0) = 0\). Substitute \(x = 0\) into \(F(x) = 3x + C\):\[F(0) = 3(0) + C = 0\]This simplifies to \(C = 0\).
5Step 5: Determine the Specific Antiderivative
Using \(C = 0\), the specific antiderivative is \(F(x) = 3x\).
6Step 6: Check for Uniqueness
Since applying the initial condition \(F(0) = 0\) led to a unique value of \(C\), in this problem the antiderivative \(F(x) = 3x\) is unique.
Key Concepts
Initial ConditionsConstant FunctionsDerivative
Initial Conditions
When solving problems involving antiderivatives, initial conditions play a crucial role. They provide specific information that helps determine the particular solution from a family of functions.
Initially, when you find an antiderivative, it includes a constant term, usually represented by "+ C". This means there are infinitely many solutions that differ only by a constant. However, an initial condition fixes the value of this constant, providing a unique function.
For instance, in our exercise, we started with the general antiderivative of the constant function \( f(x) = 3 \), which is \( F(x) = 3x + C \). The initial condition given is \( F(0) = 0 \). By substituting \( x = 0 \) into the general form, we determined that \( C \) must be zero. Thus, the specific solution became \( F(x) = 3x \), ensuring it satisfies both the derivative condition and the initial condition.
Initially, when you find an antiderivative, it includes a constant term, usually represented by "+ C". This means there are infinitely many solutions that differ only by a constant. However, an initial condition fixes the value of this constant, providing a unique function.
For instance, in our exercise, we started with the general antiderivative of the constant function \( f(x) = 3 \), which is \( F(x) = 3x + C \). The initial condition given is \( F(0) = 0 \). By substituting \( x = 0 \) into the general form, we determined that \( C \) must be zero. Thus, the specific solution became \( F(x) = 3x \), ensuring it satisfies both the derivative condition and the initial condition.
Constant Functions
A constant function, like the one in this exercise \( f(x) = 3 \), is simple yet critical for understanding antiderivatives.
For any constant function \( c \), the antiderivative is of the form \( F(x) = cx + C \), where \( C \) is an arbitrary constant.
This characteristic arises because the derivative of a linear function \( cx + C \) is \( c \), as the derivative of a constant \( C \) is zero. Thus, integrating the constant function \( c \) yields a family of linear functions. That's why constant functions play a fundamental role in calculus; they help establish the linearity in antiderivatives.
For any constant function \( c \), the antiderivative is of the form \( F(x) = cx + C \), where \( C \) is an arbitrary constant.
This characteristic arises because the derivative of a linear function \( cx + C \) is \( c \), as the derivative of a constant \( C \) is zero. Thus, integrating the constant function \( c \) yields a family of linear functions. That's why constant functions play a fundamental role in calculus; they help establish the linearity in antiderivatives.
Derivative
The derivative concept is essential for solving antiderivative problems. It’s like working backward in a puzzle.
To find an antiderivative of a function, you need to determine what function, when differentiated, gives you the original function. For example, the derivative of \( F(x) = 3x + C \) is \( f(x) = 3 \), as differentiating the \( 3x \) term results in \( 3 \), and the constant \( C \) disappears.
This relationship between a function and its derivative underpins the entire process of finding antiderivatives. Understanding this can help anyone solve differential equations or interpret initial conditions effectively. Thus, knowing how derivatives operate is key to grasping broader calculus concepts.
To find an antiderivative of a function, you need to determine what function, when differentiated, gives you the original function. For example, the derivative of \( F(x) = 3x + C \) is \( f(x) = 3 \), as differentiating the \( 3x \) term results in \( 3 \), and the constant \( C \) disappears.
This relationship between a function and its derivative underpins the entire process of finding antiderivatives. Understanding this can help anyone solve differential equations or interpret initial conditions effectively. Thus, knowing how derivatives operate is key to grasping broader calculus concepts.
Other exercises in this chapter
Problem 27
The concentration, \(C\), in \(\mathrm{ng} / \mathrm{ml}\), of a drug in the blood as a function of the time, \(t\), in hours since the drug was administered is
View solution Problem 27
Write the definite integral for the area under the graph of \(f(x)=6 x^{2}+1\) between \(x=0\) and \(x=2\). Use the Fundamental Theorem of Calculus to evaluate
View solution Problem 27
Find the integrals in problems. Check your answers by differentiation. $$ \int x \sin \left(4 x^{2}\right) d x $$
View solution Problem 28
During a surge in the demand for electricity, the rate, \(r\), at which energy is used can be approximated by $$ r=t e^{-a t} $$ where \(t\) is the time in hour
View solution