Problem 89 - Applied Calculus

Question

Find an antiderivative \(F(x)\) with \(F^{\prime}(x)=f(x)\) and \(F(0)=5\) $$f(x)=8 \sin (2 x)$$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = -4 \cos(2x) + 9 \).

Step 1: Identify the Function to Integrate

The given function to integrate is \( f(x) = 8 \sin(2x) \). We are tasked with finding its antiderivative \( F(x) \).

Step 2: Use Basic Integration Rules

We know that the integral of \( \sin(ax) \) is \( -\frac{1}{a}\cos(ax) \). Therefore, for \( f(x) = 8\sin(2x) \), we apply this rule.

Step 3: Integrate the Function

Integrate \( f(x) = 8\sin(2x) \):\[ F(x) = \int 8 \sin(2x) \, dx = 8 \cdot \left(-\frac{1}{2} \cos(2x)\right) + C = -4 \cos(2x) + C \]Here, \( C \) is the constant of integration.

Step 4: Use Initial Condition to Find C

The problem states \( F(0) = 5 \). Substitute \( x = 0 \) into the integrated function:\[ F(0) = -4 \cos(2 \cdot 0) + C = 5 \]\( \cos(0) = 1 \), so:\[ -4(1) + C = 5 \]\[ C = 5 + 4 = 9 \]

Step 5: Write the Final Antiderivative

Combine the function from Step 3 and the constant \( C \) from Step 4:\[ F(x) = -4 \cos(2x) + 9 \]

Key Concepts

Antiderivative

The antiderivative, also known as the indefinite integral, is a function whose derivative is the given function. In the context of our exercise, we are required to find the antiderivative of the function \( f(x) = 8 \sin(2x) \). This means we need to determine a function \( F(x) \) such that \( F'(x) = f(x) \).
To achieve this, we perform an integration operation on \( f(x) \). The process of finding an antiderivative involves reversing differentiation. It is important to remember that antiderivatives are not unique because the derivative of a constant is zero. As a result, the antiderivative will include a constant of integration, denoted as \( C \). This constant can be determined if an initial condition is provided, as in this exercise.

Integration

Integration is the mathematical process of finding the antiderivative of a function. It is essentially the reverse of differentiation and is used to accumulate values, calculate areas under curves, and more.
For the function \( f(x) = 8 \sin(2x) \), integration involves applying the integration rule for \( \sin(ax) \), which states that the antiderivative is \( -\frac{1}{a} \cos(ax) \). With this rule, we can integrate to find:

• \( F(x) = \int 8 \sin(2x) \, dx = 8 \cdot \left(-\frac{1}{2} \cos(2x)\right) + C = -4 \cos(2x) + C \)

This integration step results in a general solution containing a constant \( C \), which reflects the family of all possible antiderivatives. It is crucial to apply these rules correctly to derive the right form of the antiderivative.

Initial Condition

An initial condition is an extra piece of information that helps us find the specific antiderivative among the family of solutions. In our problem, the initial condition given is \( F(0) = 5 \). This information enables us to determine the constant \( C \) in our antiderivative equation.
To use the initial condition, substitute \( x = 0 \) into the antiderivative \( F(x) = -4 \cos(2x) + C \). This substitution gives us:

• \( -4 \cos(0) + C = 5 \)

Since \( \cos(0) = 1 \), we have:

• \( -4(1) + C = 5 \)

Solving for \( C \) yields \( C = 9 \). With the determined constant, the exact form of our antiderivative is \( F(x) = -4 \cos(2x) + 9 \).
This process of solving for \( C \) using the initial condition is crucial for pinpointing the unique solution that satisfies both the derivative and initial conditions.

Trig Functions

Trigonometric functions, or trig functions, like \( \sin(x) \) and \( \cos(x) \), are fundamental in calculus, especially in integration and differentiation tasks. These functions are periodic and exhibit unique properties that are useful for evaluating integrals.
In our problem, we integrated the sine function \( 8 \sin(2x) \). The integration of sinusoidal functions often involves other trigonometric functions, such as cosine. The integral of \( \sin(ax) \) is \( -\frac{1}{a}\cos(ax) \), so we used this to find the antiderivative:

• \( -\frac{1}{2} \cos(2x) \), after integrating \( \sin(2x) \)

The trigonometric identity \( \cos(0) = 1 \) was also leveraged when calculating the constant using the initial condition.
Understanding these basic trig identities and how they interact during integration is essential for successfully solving calculus problems involving these functions.