Problem 63 - Applied Calculus

Question

Find an antiderivative \(F(x)\) with \(F^{\prime}(x)=f(x)\) and \(F(0)=5\). $$ f(x)=6 x-5 $$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = 3x^2 - 5x + 5 \).

Step 1: Understand the Problem

We need to find an antiderivative \( F(x) \) such that the derivative \( F'(x) = f(x) \) is known to be \( f(x) = 6x - 5 \). Furthermore, it should satisfy the initial condition \( F(0) = 5 \).

Step 2: Find the General Antiderivative

To find the general antiderivative, integrate the function \( f(x) = 6x - 5 \). This is done by applying the power rule for integration: \[ \int (6x - 5) \, dx = \int 6x \, dx - \int 5 \, dx \].

Step 3: Integrate Each Term

Start by integrating each term separately. The integral of \( 6x \, dx \) is \( 3x^2 \) because \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \). Therefore, \( \int 6x \, dx = 6 \cdot \frac{x^{1+1}}{1+1} = 3x^2 \). The integral of \( 5 \, dx \) is \( 5x \).

Step 4: Combine the Terms

Now combine the integrated terms to find the general antiderivative: \( F(x) = 3x^2 - 5x + C \), where \( C \) is the constant of integration.

Step 5: Apply the Initial Condition

Use the given initial condition \( F(0) = 5 \) to find \( C \). Substitute \( x = 0 \) into the equation \( F(x) = 3x^2 - 5x + C \) to get: \[ F(0) = 3 \cdot 0^2 - 5 \cdot 0 + C = 5 \]. This simplifies to \( C = 5 \).

Step 6: Write the Particular Antiderivative

Substitute \( C = 5 \) into the expression for the antiderivative to find \( F(x) \): \[ F(x) = 3x^2 - 5x + 5 \]. This is the required antiderivative that satisfies \( F'(x) = f(x) \) and \( F(0) = 5 \).

Key Concepts

Antiderivative

The term **antiderivative** refers to a function that reverses the process of differentiation. Put simply, if you have a function, the antiderivative is a new function whose derivative returns you to the original function. For example, if you start with a function \( f(x) \), finding its antiderivative \( F(x) \) involves determining a function such that when you differentiate \( F(x) \), you obtain \( f(x) \).
In our example, the function \( f(x) = 6x - 5 \) requires finding an antiderivative \( F(x) \) such that \( F'(x) = f(x) \). The antiderivative is not unique because you can add a constant to any solution, resulting in what we call the "family of antiderivatives." Why is this?

• It stems from the fundamental concept that the derivative of a constant is zero, leaving room for various constants added to our solution.

This means the general antiderivative \( F(x) = 3x^2 - 5x + C \) can take an infinite number of forms, one for each possible value of \( C \). Knowing this, let's learn how to find it.

Integration

**Integration** is the process we use to find the antiderivative of a function. It's like piecing together the original image from the puzzle of its derivative. There are many techniques for integration, but one commonly used in simple problems is the **power rule for integration**.
For our case, the function \( f(x) = 6x - 5 \) was integrated to find \( F(x) \). The integration process involved breaking down the problem:

• Integrate each term separately: \( \int 6x \, dx \) and \( \int -5 \, dx \).
• The results, when combined, yield the general form \( F(x) = 3x^2 - 5x + C \).

The integration is slightly trickier because it deals with two components: \( 6x \) and \( -5 \). After integrating each, the results were combined to reveal the general antiderivative before accounting for any constants of integration.

Initial condition

An **initial condition** helps us find the specific antiderivative from the 'family' of possible solutions. Think of it as a key that locks in the exact solution we need by pinpointing the appropriate constant \( C \). In problems involving antiderivatives, you're often given an initial condition.
In our example, the initial condition was \( F(0) = 5 \). Applying this involves substituting \( x = 0 \) into the general solution \( F(x) = 3x^2 - 5x + C \) and setting it equal to 5:

• This yielded \( C = 5 \) after solving the equation \( 3 \cdot 0^2 - 5 \cdot 0 + C = 5 \).

With the value of \( C \) determined, you're ready to write the particular antiderivative which uniquely corresponds to the given condition, completing the picture as shown by \( F(x) = 3x^2 - 5x + 5 \).

Power rule

When dealing with polynomials, the **power rule** is a fundamental tool in integration. It provides a handy shortcut for simple computations, especially when functions take the form \( x^n \). When applying the power rule, the idea is straightforward:

• For a term like \( x^n \), the integral is \( \frac{x^{n+1}}{n+1} \) plus a constant \( C \).

Using our function \( f(x) = 6x - 5 \), we break it down into its components:

• For \( 6x \), apply the power rule to get \( 3x^2 \) because \( 6 \cdot \frac{x^{1+1}}{1+1} = 3x^2 \).
• For \( -5 \), integrate to get \( -5x \).

The power rule is immensely helpful because it simplifies integration through rapid adjustments of exponents and coefficients, offering a neat solution method for a wide array of polynomial structures.