Problem 45

Question

Find an antiderivative \(F(x)\) with \(F^{\prime}(x)=\) \(f(x)\) and \(F(0)=0 .\) Is there only one possible solution? $$f(x)=2+4 x+5 x^{2}$$

Step-by-Step Solution

Verified
Answer
The antiderivative is \( F(x) = 2x + 2x^2 + \frac{5x^3}{3} \), and it is unique.
1Step 1: Identify the Function to Integrate
The function given is \( f(x) = 2 + 4x + 5x^2 \). We need to find the antiderivative, which means we need to integrate this function term by term.
2Step 2: Integrate Term by Term
To find the antiderivative, integrate each term of \( f(x) \):- The integral of \( 2 \) is \( 2x \).- The integral of \( 4x \) is \( \frac{4x^2}{2} = 2x^2 \).- The integral of \( 5x^2 \) is \( \frac{5x^3}{3} \).
3Step 3: Write the General Antiderivative
Combine the integrated terms to write the general form of the antiderivative:\( F(x) = 2x + 2x^2 + \frac{5x^3}{3} + C \), where \( C \) is the constant of integration.
4Step 4: Apply the Initial Condition
We're given that \( F(0) = 0 \). Substitute \( x = 0 \) in the antiderivative to find \( C \):\( F(0) = 2(0) + 2(0)^2 + \frac{5(0)^3}{3} + C = 0 \). This simplifies to \( C = 0 \).
5Step 5: Write the Particular Antiderivative
Substitute \( C = 0 \) back into the antiderivative:\( F(x) = 2x + 2x^2 + \frac{5x^3}{3} \).
6Step 6: Determine Uniqueness of the Antiderivative
Since we derived \( C = 0 \) from the given condition, there is only one possible antiderivative that satisfies \( F(0) = 0 \).

Key Concepts

IntegrationInitial ConditionConstant of Integration
Integration
Integration is a key concept in calculus that involves reversing the process of differentiation. If you have a function that you know is the derivative of another function, integration helps you find the original function. For the exercise given, the goal is to find the function whose derivative is the provided function \( f(x) = 2 + 4x + 5x^2 \).
When integrating, we break down each term to find their antiderivatives. For instance, the antiderivative of a constant, like 2, is simply the constant times the variable, resulting in \( 2x \). For powers of \( x \), like \( 4x \) and \( 5x^2 \), the power rule of integration is applied: increase the exponent by one and then divide by the new exponent. This gives us antiderivatives of \( 2x^2 \) and \( \frac{5x^3}{3} \) respectively.
Integration brings us to a general form, which always includes a constant of integration.
Initial Condition
An initial condition is a piece of information that provides the specific value of the function at a certain point. In integration, after finding the general antiderivative, this condition is used to find the precise function, also known as the particular solution.
In our problem, we're given \( F(0) = 0 \). This tells us exactly what the value of our antiderivative \( F(x) \) must be when \( x = 0 \).
  • This condition helps fix the value of the constant of integration.
  • Inserting \( x = 0 \) in the general formula \( F(x) = 2x + 2x^2 + \frac{5x^3}{3} + C \) will ensure accuracy.
The initial condition is crucial as it allows us to find a unique solution that satisfies the given criteria.
Constant of Integration
Every time you integrate a function, you add a constant of integration, often represented by \( C \). This constant arises because the derivative of a constant is zero, so the original function could differ by any constant without affecting the derivative.
In the presented exercise, the general form of the antiderivative was \( F(x) = 2x + 2x^2 + \frac{5x^3}{3} + C \). The constant \( C \) is refined using the initial condition. This constant is essential in determining the specific function among the infinite possibilities that match the differential equation.
  • Setting \( F(0) = 0 \) revealed that \( C \) must be 0, narrowing down our solution to one precise function.
  • The constant of integration adapts the generalized solution to a specific individual solution given the context of the problem.
The magic of \( C \) ensures that our integration process is versatile enough to accommodate various real-world scenarios and applications.