Problem 37
Question
In each of Exercises \(35-40\), an expression \(f(x)\) is given. Find all functions \(F\) such that \(F^{\prime}(x)=f(x)\). $$ x^{2}+\pi $$
Step-by-Step Solution
Verified Answer
The function is \( F(x) = \frac{x^3}{3} + \pi x + C \).
1Step 1: Understanding the Problem Statement
We are given the function \( f(x) = x^2 + \pi \). We need to find a function \( F(x) \) such that when differentiated, it equals \( f(x) \). In other words, we are finding the antiderivative or indefinite integral of \( f(x) \).
2Step 2: Set Up the Indefinite Integral
To find the function \( F(x) \), we need to compute the indefinite integral: \[ F(x) = \int (x^2 + \pi) \, dx \] This will help us find all functions whose derivative is \( f(x) \).
3Step 3: Break Down the Integral
The integral can be broken into two parts: \[ F(x) = \int x^2 \, dx + \int \pi \, dx \]This simplifies the problem by allowing us to handle each part separately.
4Step 4: Integrate Each Part Separately
First, solve \( \int x^2 \, dx \):Apply the power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] Substitute \( n = 2 \): \[ \int x^2 \, dx = \frac{x^3}{3} + C_1 \]Second, solve \( \int \pi \, dx \):Since \( \pi \) is a constant, integration is straightforward: \[ \int \pi \, dx = \pi x + C_2 \]
5Step 5: Combine the Results
Add the results of the separate integrations together:\[ F(x) = \frac{x^3}{3} + \pi x + C \]Where \( C = C_1 + C_2 \) is the constant of integration, representing the family of all functions whose derivative is \( f(x) \).
Key Concepts
AntiderivativePower Rule for IntegrationConstant of IntegrationDifferentiation and Integration
Antiderivative
Finding an antiderivative means determining a function whose derivative matches a given function. In the context of the problem, we are given the function \( f(x) = x^2 + \pi \), and are asked to find a function \( F(x) \) such that \( F'(x) = f(x) \). This process is also known as finding the indefinite integral of \( f(x) \).
An important thing to note is that the antiderivative of a function is not unique. Any number of functions can have the same derivative, differing by a constant. This highlights why the result of an indefinite integral always includes an arbitrary constant, often denoted by \( C \).
Understanding antiderivatives is fundamental to calculus as it connects the process of differentiation with integration, essentially reversing the operation of finding the derivative.
An important thing to note is that the antiderivative of a function is not unique. Any number of functions can have the same derivative, differing by a constant. This highlights why the result of an indefinite integral always includes an arbitrary constant, often denoted by \( C \).
Understanding antiderivatives is fundamental to calculus as it connects the process of differentiation with integration, essentially reversing the operation of finding the derivative.
Power Rule for Integration
The power rule for integration is a simple and essential technique for finding the antiderivative of functions in the form \( x^n \). The rule states that \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]provided that \( n eq -1 \).
For our exercise, we can apply the power rule to the term \( x^2 \) to find its antiderivative. By setting \( n = 2 \), we substitute into the formula:
- \( \int x^2 \, dx = \frac{x^3}{3} + C_1 \)
This illustrates how simple it is to handle polynomials using this rule, making it an invaluable tool for solving integrals.
For our exercise, we can apply the power rule to the term \( x^2 \) to find its antiderivative. By setting \( n = 2 \), we substitute into the formula:
- \( \int x^2 \, dx = \frac{x^3}{3} + C_1 \)
This illustrates how simple it is to handle polynomials using this rule, making it an invaluable tool for solving integrals.
Constant of Integration
When finding an antiderivative, it is crucial to remember the constant of integration, denoted by \( C \). The constant \( C \) represents an arbitrary constant value which can shift the graph vertically in the case of polynomial functions like in our example.
This is because differentiating a constant yields zero, and thus any constant value added to an antiderivative will not change the differentiation result. In the solution, the statement \( C = C_1 + C_2 \) signifies the aggregated constant term from integrating different parts.
In practice, the constant of integration ensures that every possible antiderivative is accounted for, creating a whole family of functions that share \( f(x) \) as their derivative.
This is because differentiating a constant yields zero, and thus any constant value added to an antiderivative will not change the differentiation result. In the solution, the statement \( C = C_1 + C_2 \) signifies the aggregated constant term from integrating different parts.
In practice, the constant of integration ensures that every possible antiderivative is accounted for, creating a whole family of functions that share \( f(x) \) as their derivative.
Differentiation and Integration
Differentiation and integration are core operations of calculus, often viewed as inverse processes. Differentiation involves finding a derivative, which tells us the rate of change of a function. In contrast, integration is about finding the original function given its rate of change.
Understanding this inverse relationship is vital when analyzing functions. In the given exercise, you need to integrate \( f(x) = x^2 + \pi \) to discover the function \( F(x) \) from which it was derived. This requires reversing the differentiation process to reveal the integral and thus the potential shapes and shifts of \( F(x) \).
With differentiation and integration, calculus offers powerful tools to model and solve a broad range of real-world problems, from plotting a journey's progress to understanding economic trends.
Understanding this inverse relationship is vital when analyzing functions. In the given exercise, you need to integrate \( f(x) = x^2 + \pi \) to discover the function \( F(x) \) from which it was derived. This requires reversing the differentiation process to reveal the integral and thus the potential shapes and shifts of \( F(x) \).
With differentiation and integration, calculus offers powerful tools to model and solve a broad range of real-world problems, from plotting a journey's progress to understanding economic trends.
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