Problem 21 - Calculus Single Variable

Question

Compute \(F(c)\) from the given information. $$ F^{\prime}(x)=6 x^{2}, F(1)=3, c=0 $$

Step-by-Step Solution

Answer

\( F(c) = 1 \) when \( c = 0 \).

Step 1: Find the Antiderivative

To find the function \( F(x) \), we need to compute the antiderivative of \( F'(x) = 6x^2 \). The antiderivative of \( 6x^2 \) is found by increasing the exponent of \( x \) by 1 and then dividing by that new exponent: \[ F(x) = \frac{6x^3}{3} + C = 2x^3 + C \].

Step 2: Use the Initial Condition to Find C

We know from the problem statement that \( F(1) = 3 \). Substitute \( x = 1 \) into the function to find \( C \): \[ F(1) = 2(1)^3 + C = 3 \]. Thus, \( 2 + C = 3 \), which means \( C = 1 \).

Step 3: Write the Complete Function

With the constant \( C = 1 \), we can now write the complete function: \[ F(x) = 2x^3 + 1 \].

Step 4: Compute F(c)

Now that we have the complete function \( F(x) = 2x^3 + 1 \), we can find \( F(c) \) where \( c = 0 \). Substitute \( x = 0 \) into the function: \[ F(0) = 2(0)^3 + 1 = 1 \].

Key Concepts

Understanding Antiderivatives

In calculus, the concept of antiderivatives provides a foundation for understanding how functions accumulate change. The process of finding an antiderivative is known as integration, which essentially works in reverse of differentiation.
When we have a derivative, for instance, \( F'(x) = 6x^2 \), finding the antiderivative involves restoring the original function. This is done by reversing the differentiation process.
To find the antiderivative of \( 6x^2 \), increase the exponent of \( x \) by 1, making it \( x^3 \), and then divide the entire term by this new exponent. This process results in \( \frac{6x^3}{3} \), simplifying to \( 2x^3 \).

• Antiderivatives are not unique. They include an arbitrary constant \( C \).
• Integrating a function provides a family of functions represented as \( F(x) = 2x^3 + C \).

Understanding and computing antiderivatives are crucial skills for solving problems related to infinite sequences of changes.

The Role of Initial Conditions

An initial condition is a necessary piece of information for determining the unique solution from a family of antiderivatives. In our problem, the initial condition is given by \( F(1) = 3 \). This data point helps identify the specific antiderivative relevant to the problem.
When we substitute \( x = 1 \) into the antiderivative formula \( F(x) = 2x^3 + C \), we get \( F(1) = 2(1)^3 + C \). This simplifies to \( 2 + C \), which must equal 3 based on the initial condition.
Therefore, by solving \( 2 + C = 3 \), we find that \( C = 1 \). This concrete value for \( C \) turns our family of potential solutions into a unique function:

• Initial conditions point out the specific solution.
• They convert a general solution into a specific one for real-world applications.

Knowing how to use initial conditions is a key computational step in many calculus problems.

Using the Integration Constant

The integration constant, denoted by \( C \), is vital when working with indefinite integrals. It represents the infinite number of vertical shifts possible for any antiderivative.
When we integrate \( F'(x) = 6x^2 \), our antiderivative includes this constant, forming \( F(x) = 2x^3 + C \). Initially, \( C \) is unknown because indefinite integrals do not contain specific limits of integration.
By applying the initial condition, the exact value of \( C \) is uncovered, ultimately ensuring a unique solution that fits the problem's constraints.

• The integration constant adjusts for shifts along the y-axis in the function's graph.
• Without knowing \( C \), we cannot definitively describe the function's position.

Understanding the role of \( C \) empowers problem solvers to handle broader classes of problems, converting general solutions into specific ones that match particular scenarios.

Polynomial Functions in Calculus

Polynomial functions are some of the most basic yet essential structures in calculus, appearing frequently in the study of derivatives and integrals. A polynomial function is composed of terms like \( ax^n \) where \( n \) is a non-negative integer.
In our problem, we work with a polynomial derivative \( F'(x) = 6x^2 \). The antiderivative, \( F(x) = 2x^3 + 1 \), is also a polynomial, showcasing the differentiation and integration processes that are simpler and more intuitive when applied to polynomials.

• Polynomials are easily differentiated and integrated term by term.
• They serve as excellent approximations of more complex functions.

Recognizing and working with polynomial functions supports broader understanding in calculus, forming a bridge between basic mathematical concepts and more sophisticated calculus techniques.