Question

General results Evaluate the following integrals in which the function \(f\) is unspecified. Note that \(f^{(p)}\) is the pth derivative of \(f\) and \(f^{p}\) is the pth power of \(f .\) Assume \(f\) and its derivatives are continuous for all real numbers. \(\int\left(f^{(p)}(x)\right)^{n} f^{(p+1)}(x) d x,\) where \(p\) is a positive integer, \(n \neq-1\)

Short answer

Question: Evaluate the integral \(\int\left(f^{(p)}(x)\right)^{n} f^{(p+1)}(x) \, dx\), where \(f^{(p)}(x)\) is the pth derivative of an unspecified function \(f\). Answer: The evaluated integral is \(\frac{\left(f^{(p)}(x)\right)^{n+1}}{n+1} + C\), where C is the integration constant.

Step-by-step solution

Rewrite the integral

The given integral can be written as: \(\int\left(f^{(p)}(x)\right)^{n} f^{(p+1)}(x) \, dx\)

Use integration by substitution

To solve the integral, let \(u = f^{(p)}(x)\); then, we have \(u^n\) in the integrand. We will also need to find the derivative of \(u\), i.e., \(du/dx\). Taking the derivative on both sides, we get: \(\frac{d}{dx}(u) = \frac{d}{dx}(f^{(p)}(x))\) \(\frac{du}{dx}= f^{(p+1)}(x)\) Now, multiply both sides by \(dx\): \(du = f^{(p+1)}(x) \, dx\)

Substitute and simplify

Now we'll substitute \(u\) and \(du\) into the integral: \(\int\left(f^{(p)}(x)\right)^{n} f^{(p+1)}(x) \, dx = \int u^n \, du\)

Evaluate the integral

Now, we have a simple power function to integrate: \(\int u^n \, du = \frac{u^{n+1}}{n+1} + C\) Here, \(C\) is the integration constant.

Replace the substitution

We must substitute back so that we have the result in terms of \(x\). Recall that \(u = f^{(p)}(x)\): \(\frac{u^{n+1}}{n+1} + C = \frac{\left(f^{(p)}(x)\right)^{n+1}}{n+1} + C\)

Write the final result

The final result is the evaluated integral in terms of \(x\): \(\int\left(f^{(p)}(x)\right)^{n} f^{(p+1)}(x) \, dx =\frac{\left(f^{(p)}(x)\right)^{n+1}}{n+1} + C\)

Key concepts

Integration by Substitution

Understanding integration by substitution is crucial to simplifying complex integrals into easier forms. It involves selecting a part of the integrand to be a new variable, say \( u \). This new variable helps reframe the integral into a simpler form.

Here's how it generally works:

• Identify a function inside the integral, which when differentiated, resembles another part of the integrand.
• Make a substitution, \( u = g(x) \), where \( g(x) \) is your chosen function, and find the derivative, \( du/dx \).
• Multiply both sides by \( dx \) to express \( du \) in terms of \( dx \): \( du = g'(x) \, dx \).

You then replace the original variables in the integral with your new \( u \) and \( du \). This transforms the integral into a simpler function of \( u \), which is easier to evaluate. Like in the exercise, we used \( u = f^{(p)}(x) \) and found that \( du = f^{(p+1)}(x) \, dx \), leading to an easier integral \( \int u^n \, du \).

Integration by substitution is akin to the reverse process of differentiation known as the chain rule, allowing you to backtrace a derivative to its antiderivative.

Derivatives

Derivatives are fundamental to understanding calculus and involve the rate of change of a function with respect to a variable. Consider a function \( f(x) \), the derivative \( f'(x) \) provides the slope or speed of change at any point \( x \).

Key points about derivatives:

• They are notations such as \( f'(x) \), \( \frac{df}{dx} \), or \( Df(x) \).
• Basic rules include the power rule, product rule, quotient rule, and chain rule.
• Higher-order derivatives like \( f''(x) \), \( f^{(3)}(x) \), etc., represent the rate of change of the previous derivative.

In the context of our exercise, the pth derivative \( f^{(p)}(x) \) and its next derivative \( f^{(p+1)}(x) \) were used to form the integral. The process involved understanding how these derivatives relate via integration by substitution. Remember, for continuous functions, these derivatives provide a smooth and predictable change, critical in simplifying integrals and other calculus operations.

Continuous Functions

Continuous functions are mathematical functions with no breaks, jumps, or holes in their graphs. They allow calculus operations like differentiation and integration to be performed smoothly. If a function is continuous over a specific interval, it means you can draw the graph without lifting the pencil.

Here’s what you should know:

• A function \( f(x) \) is continuous at \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \).
• Continuity across all real numbers means \( f \) is seamless, enabling consistent application of calculus techniques.
• If a function is not continuous, its derivatives or integrals may not exist in the problematic intervals.

In the given exercise, assuming \( f \) and \( f^{(p)} \) are continuous ensures that the integral can be formed and evaluated properly. Without continuity, unexpected behavior in the function can lead to incorrect or undefined calculations. Being well-versed with continuous functions simplifies the process of leveraging integration and derivatives effectively in various mathematical problems.