Problem 51 - Calculus and its applications

Question

Find f such that: $$ f^{\prime}(x)=5 x^{2}+3 x-7, \quad f(0)=9 $$

Step-by-Step Solution

Answer

The function is \( f(x) = \frac{5}{3}x^3 + \frac{3}{2}x^2 - 7x + 9 \).

Step 1: Identify the Task

We need to find the function \( f(x) \) given its derivative \( f'(x) = 5x^2 + 3x - 7 \) and the initial condition \( f(0) = 9 \). This process involves solving for the antiderivative and using the initial condition to find the constant of integration.

Step 2: Integrate the Derivative Function

Find the antiderivative of \( f'(x) = 5x^2 + 3x - 7 \). Perform the integration term by term:\[\int (5x^2 + 3x - 7) \, dx = \int 5x^2 \, dx + \int 3x \, dx - \int 7 \, dx.\]This results in:\[f(x) = \frac{5}{3}x^3 + \frac{3}{2}x^2 - 7x + C.\]

Step 3: Use the Initial Condition to Find \( C \)

Substitute \( x = 0 \) into the integrated function and set it equal to 9:\[f(0) = \frac{5}{3}(0)^3 + \frac{3}{2}(0)^2 - 7(0) + C = 9.\]Thus, we have \( C = 9 \).

Step 4: Write the Final Function

Substitute the value of \( C \) back into the expression found in Step 2:\[f(x) = \frac{5}{3}x^3 + \frac{3}{2}x^2 - 7x + 9.\]

Key Concepts

Antiderivatives

In calculus, finding an antiderivative of a function involves determining the original function from its derivative. This process is the reverse of differentiation. When you are given a derivative, such as \( f'(x) = 5x^2 + 3x - 7 \), you want to find a function \( f(x) \) whose derivative is the given function. This is known as finding the antiderivative or the indefinite integral.

An antiderivative of a function \( g(x) \) is any function \( F(x) \) such that \( F'(x) = g(x) \). In simple terms, by integrating \( f'(x) \), you can rectify it to its main form \( f(x) \). The general solution to an indefinite integral includes a constant of integration, \( C \), because the derivative of a constant is zero. Thus, different constants will result in different antiderivatives for the same function, and they will all differ by a constant amount.

Initial Conditions

An initial condition is a specific piece of information that helps in finding the particular solution to a differential equation or an integral equation. It is a value that defines some aspect of the function at a specific point, allowing us to solve for the constant of integration \( C \). In this exercise, the initial condition given is \( f(0) = 9 \).

When we identify the antiderivative of a function, it includes a constant \( C \). To find the specific solution to fit our original problem, we need to use the initial condition. This step practically involves substituting the specific \( x \) and \( f(x) \) given by the initial condition into the antiderivative, then solving for \( C \).

• The antiderivative is generalized as \( f(x) = \frac{5}{3}x^3 + \frac{3}{2}x^2 - 7x + C \).
• Using the initial condition \( f(0) = 9 \), we substitute into this antiderivative to find \( C \).
• This tells us exactly where this curve begins on the graph, establishing the particular form of the function.

Integration Techniques

Integration techniques are methods used to find an antiderivative or the integral of a derivative function. A critical aspect of calculus, integration allows you to solve problems that involve area under a curve, among other applications.

The primary technique demonstrated in this exercise is **term-by-term integration**. Given the derivative function \( f'(x) = 5x^2 + 3x - 7 \), we integrate each term individually.

• The integral of \( 5x^2 \) is \( \frac{5}{3}x^3 \).
• The integral of \( 3x \) is \( \frac{3}{2}x^2 \).
• The integral of \(-7\) is \(-7x \).

This technique involves integrating each term separately and then combining the results, along with the constant of integration \( C \), to form the overall integral. This results in the expression \( f(x) = \frac{5}{3}x^3 + \frac{3}{2}x^2 - 7x + C \).

Another method, not used here but crucial in other contexts, is **substitution**. Substitution is particularly useful for more complex expressions or when the direct term-by-term approach is cumbersome. Understanding the right integration technique is key in handling complex calculus problems efficiently.