Problem 65 - 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: Understand the Problem

We are given the derivative \( F'(x) = f(x) = 8 \sin(2x) \). We need to find the antiderivative \( F(x) \) such that \( F(0) = 5 \).

Step 2: Find the General Antiderivative

The antiderivative of \( \sin(2x) \) is \( -\frac{1}{2}\cos(2x) \). Therefore, the antiderivative of \( 8 \sin(2x) \) is \( -4 \cos(2x) + C \), where \( C \) is the constant of integration. Thus, \( F(x) = -4 \cos(2x) + C \).

Step 3: Use the Initial Condition

We are given \( F(0) = 5 \). Substitute \( x = 0 \) into \( F(x) = -4 \cos(2x) + C \) to find \( C \). \[ F(0) = -4 \cos(0) + C = 5\] Since \( \cos(0) = 1 \), the equation becomes: \(-4 \times 1 + C = 5\), which results in \( C = 9 \).

Step 4: State the Specific Antiderivative

Substitute \( C = 9 \) back into \( F(x) = -4 \cos(2x) + C \) to give the specific antiderivative: \( F(x) = -4 \cos(2x) + 9 \).

Key Concepts

Trigonometric Functions

Trigonometric functions are an essential aspect of calculus, and understanding them is crucial when dealing with derivatives and antiderivatives. In our exercise, the function we deal with is trigonometric:

• The original function given is \( f(x) = 8 \sin(2x) \).
• \( \sin(2x) \) is a sine function, which is one of six basic trigonometric functions.
• Trigonometric functions are periodic, meaning they repeat their values in regular intervals. For \( \sin(x) \), this interval is \(2\pi\).
• The function \( \sin(kx) \) describes a sine wave, where \( k \) is a constant that affects the frequency of the function.

Understanding these properties helps to recognize how the antiderivative of a sine function involves a cosine function, since integrating \( \sin(x) \) results in \( -\cos(x) \). This integration follows the basic rule of integrating trigonometric functions, which transforms derivatives back into their original function forms, considering the negative sign for \( \sin(x) \). To know it fully, base reinforcement in both differentiation and integration processes of basic trigonometric functions is needed, especially memorizing the integration of \( \sin(x) \), \( \cos(x) \), and other counterparts.

Constant of Integration

When finding antiderivatives, the constant of integration plays a critical role.

• An antiderivative of a function \( f(x) \) is not unique; instead, you obtain a family of functions because of this constant \( C \).
• The general form of an antiderivative for any function will include \( C \) because differentiating a constant results in zero, thus having no effect on the derivative.

In our exercise, once we determine that the antiderivative of \( f(x) = 8 \sin(2x) \) is \( F(x) = -4 \cos(2x) + C \), the presence of \( C \) reflects all possible vertical shifts of the curve described by the function. Since any constant derivative is zero, retaining \( C \) ensures that every function’s potential original shape is captured. This crucial constant allows us to adjust the antiderivative to meet specific conditions, which are often initial conditions provided as part of the problem. Understanding the purpose of the constant of integration enhances the grasp of how functions can represent various solutions.

Initial Conditions

Initial conditions are the specific values given in a problem to find a precise and unique antiderivative or solution.

• They function as anchor points, fixing the state of a function at a particular point.
• In the problem at hand, the initial condition is \( F(0) = 5 \).
• Applying the initial condition entails substituting the given point into your general antiderivative function to solve for the constant of integration \( C \).

In the example, substituting our initial condition \( F(0) = 5 \) into \( F(x) = -4 \cos(2x) + C \), we determine \( C = 9 \). This substitution considers the boundary condition to derive \( F(x) \) as \( F(x) = -4 \cos(2x) + 9 \). Such initial conditions ensure that the function not just solves the differential equation but also adheres to specific values, securing an accurate model that fits the scenario or real-life application precisely.