Problem 48 - Calculus Single Variable

Question

In each of Exercises \(41-48\), use the given information to find \(F(c)\). $$ F^{\prime}(x)=2^{x} \cdot \ln ^{2}(2), \quad F(1)=1+2 \ln (2), \quad c=3 $$

Step-by-Step Solution

Answer

\( F(3) = 8 \cdot \ln(2) + 1 \).

Step 1: Identify the Problem

We need to find the value of the function \( F(c) \), which means evaluating the function \( F \) at \( c = 3 \). We will use the given derivative \( F'(x) \) and initial condition \( F(1) = 1 + 2 \ln(2) \).

Step 2: Find the Antiderivative

To find \( F(x) \), we need to integrate the derivative \( F'(x) = 2^x \cdot \ln^2(2) \). This requires recognizing that \( \ln^2(2) \) is a constant, so we integrate as follows: \[ \int 2^x \cdot \ln^2(2) \, dx = \ln^2(2) \cdot \int 2^x \, dx = \ln^2(2) \cdot \frac{2^x}{\ln(2)} + C, \]where \( C \) is the constant of integration.

Step 3: Simplify the Integrated Function

After integrating, the antiderivative of \( F'(x) \) is \[ F(x) = \frac{2^x}{\ln(2)} \cdot \ln^2(2) + C. \]This simplifies to\[ F(x) = 2^x \cdot \ln(2) + C. \]

Step 4: Use Initial Condition to Find C

We use the given initial condition \( F(1) = 1 + 2 \ln(2) \) to find \( C \). Substitute \( x = 1 \) into \( F(x) \):\[ 2^1 \cdot \ln(2) + C = 1 + 2 \ln(2). \]This simplifies to\[ 2 \ln(2) + C = 1 + 2 \ln(2). \]Thus, \( C = 1. \)

Step 5: Evaluate F(x) at c = 3

Now with \( C = 1 \), we have\[ F(x) = 2^x \cdot \ln(2) + 1. \]We plug in \( x = 3 \):\[ F(3) = 2^3 \cdot \ln(2) + 1 = 8 \cdot \ln(2) + 1. \]

Key Concepts

Antiderivative

Calculus often explores finding antiderivatives, which are critical in undoing the process of differentiation. Essentially, when you have a derivative of a function, the antiderivative represents the original function before it was differentiated. In this exercise, we're given the derivative \( F'(x) = 2^x \cdot \ln^2(2) \). Our job is to reverse this process through integration to obtain \( F(x) \), the antiderivative.

• Identify Constants: Recognize \( \ln^2(2) \) as a constant when integrating \( F'(x) \).
• Use Integration Techniques: Remember that integrating exponential functions often involves recognizing which constants and variables need special attention.

By finding the antiderivative \( F(x) \), we reveal the pattern of the original function based on its rate of change expressed by the derivative.

Initial Condition

An initial condition is like a clue that helps us find the exact form of the antiderivative. Once we find \( F(x) \) in the indefinite form \( F(x) = 2^x \cdot \ln(2) + C \), where \( C \) is the integration constant, the initial condition provides us the value \( F(1) = 1 + 2 \ln(2) \).
This condition helps identify the specific value of \( C \). By placing \( x = 1 \) in the equation, we calculate and adjust \( C \) to ensure the equation satisfies the value \( F(1) \).

• Personalize the Equation: Initial conditions personalize a general antiderivative.
• Solve for \( C \): Replace the variable with specific values to find the constant of integration, completing the antiderivative solution.

Knowing this, we find that \( C = 1 \), setting the stage for evaluating the function with complete accuracy.

Integration

Integration is the core process of finding the antiderivative. It involves accumulating values, typically area under a curve, which results in finding \( F(x) \). Think of it as the reverse operation of differentiation.
Here, the integration of \( 2^x \cdot \ln^2(2) \) highlights the exponential function and constants within the equation. We handle the constant \( \ln^2(2) \) by factoring it outside the integral, simplifying:

• Compute Similar Integrals: Recognize patterns from simpler parts, like \( \int 2^x \, dx = \frac{2^x}{\ln(2)} \).
• Include the Constant: Multiply back the constant in the simplification process, which merges into our final expression. This results in \( \frac{2^x}{\ln(2)} \cdot \ln^2(2) + C \).

Integrating in this context helps us rebuild the function \( F(x) = 2^x \cdot \ln(2) + C \) from its differential version.

Function Evaluation

After determining the full expression of \( F(x) \) using antiderivatives and the initial condition, we proceed to function evaluation. This step involves plugging specific values into our derived function to calculate precise outputs.
Our goal is to find \( F(c) \) where \( c = 3 \). Simply substitute \( x = 3 \) into our full function:

• Substitution Method: Insert \( x = 3 \) into \( F(x) = 2^x \cdot \ln(2) + 1 \) to parse the specific evaluation.
• Calculation of Constants: Recognize that constants within the function stay unchanged through these substitutions.

Finally, computing \( F(3) \), we find the result to be \( 8 \cdot \ln(2) + 1 \), which satisfies the problem requirements for determining \( F(c) \) precisely.