Problem 41 - Calculus Single Variable

Question

In each of Exercises \(41-48\), use the given information to find \(F(c)\). $$ F^{\prime}(x)=3 x^{2}, \quad F(2)=-4, \quad c=3 $$

Step-by-Step Solution

Answer

\( F(c) = 15 \)

Step 1: Understand the Problem

We need to find the value of the function \( F(c) \), given the derivative \( F'(x) = 3x^2 \), an initial condition \( F(2) = -4 \), and \( c = 3 \). This requires us to find \( F(x) \) by integrating \( F'(x) \).

Step 2: Integrate the Derivative

The given derivative is \( F'(x) = 3x^2 \). To find \( F(x) \), we need to integrate \( 3x^2 \): \[ F(x) = \int 3x^2 \, dx \]The integral of \( 3x^2 \) is \( x^3 \) plus a constant \( C \), so:\[ F(x) = x^3 + C \]

Step 3: Apply the Initial Condition

We use the initial condition \( F(2) = -4 \) to find the constant \( C \). Substitute \( x = 2 \) into the equation:\[ F(2) = 2^3 + C = -4 \]\[ 8 + C = -4 \]Solving for \( C \), we get \( C = -12 \).

Step 4: Write the Function

Substitute the value of \( C \) back into the equation for \( F(x) \):\[ F(x) = x^3 - 12 \]

Step 5: Calculate \( F(c) \)

Now that we have \( F(x) = x^3 - 12 \), we can find \( F(c) \) by substituting \( c = 3 \):\[ F(3) = 3^3 - 12 \]\[ F(3) = 27 - 12 \]\[ F(3) = 15 \]

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that helps us find functions when their derivatives are given. The main idea is to reverse the process of differentiation. When you're given a derivative, like in our exercise with \( F'(x) = 3x^2 \), we need to integrate it to find the original function \( F(x) \). The integration process involves determining a function whose derivative matches the one we have. Integrating \( 3x^2 \) means calculating \( \int 3x^2 \, dx \). This leads to the integral \( x^3 + C \), where \( C \) is a constant. This process is like undoing the derivative, helping us "go backward" to find the original equation.

• Integrals offer ways to calculate areas under curves.
• They are essential in solving real-world problems involving accumulations, like total distances or quantities.

Understanding integral calculus gives us tools to piece together information from rates or changes to find original measurable quantities.

Initial Conditions

Initial conditions are key to solving integrals when you find an unknown constant. In calculus, when we integrate a function like \( 3x^2 \), we get \( F(x) = x^3 + C \). But without an initial condition, \( C \) could be virtually any number. In our exercise, we know \( F(2) = -4 \). This tells us exactly where the curve of such a function touches a specific point on the graph when \( x = 2 \). Applying the initial condition, we substitute \( x = 2 \) into our equation, giving \( 8 + C = -4 \). Solving for \( C \) results in \( C = -12 \).

• Initial conditions help specify our function in context, removing ambiguity.
• They allow us to solve precisely for constants, refining solutions.

Initial conditions are like the unique key that perfectly fits into the lock of a generalized solution created by integration.

Constant of Integration

The constant of integration, denoted usually as \( C \), emerges from the indefinite integral of a function. Every time you integrate, you must include this constant because derivatives "forget" constant terms. When you differentiate something with respect to \( x \), constants vanish. Thus, any function derived initially could differ by some constant yet have the same derivative. In our example, though the integration of \( 3x^2 \) gives us \( x^3 \), we write it as \( x^3 + C \) to acknowledge all potential original functions. To pinpoint the actual constant, we rely on specific initial conditions. The constant of integration accounts for all those possible translations of a curve that differentiate identically.

• Whenever integrating, remember to add \( C \) to reflect full generality.
• Solving for \( C \) personalizes the solution to specific conditions.

This constant ensures we don't overlook any original information when retracing steps from derivative functions.