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. $$\int\left(5 f^{3}(x)+7 f^{2}(x)+f(x)\right) f^{\prime}(x) d x$$

Short answer

Question: Evaluate the integral of the function: $$\int\left(5 f^{3}(x)+7 f^{2}(x)+f(x)\right) f^{\prime}(x) d x$$ Answer: The integral evaluates to $$\frac{5}{4}f^4(x) + \frac{7}{3}f^3(x) + \frac{1}{2}f^2(x) + C$$.

Step-by-step solution

Understand the integral

Given the integral: $$\int\left(5 f^{3}(x)+7 f^{2}(x)+f(x)\right) f^{\prime}(x) d x$$ The goal is to evaluate this integral with respect to \(x\). Notice that, the integral is a combination of the different powers of \(f(x)\) multiplied by its derivative \(f'(x)\).

Apply substitution

To solve this integral, we will use a substitution. Let \(u = f(x)\). Calculate the derivative of \(u\) with respect to \(x\): $$\frac{d u}{d x} = f^{\prime}(x)$$ Now, substitute \(u\) into the integral and rewrite the integral in terms of \(u\): $$\int\left(5 u^3 + 7 u^2 + u\right) d u$$

Integrate using the power rule

Now, integrate each term of the integral with respect to \(u\) using the power rule: $$\int 5u^3 du + \int 7u^2 du + \int u du$$ $$= \frac{5}{4}u^4 + \frac{7}{3}u^3 + \frac{1}{2}u^2 + C$$

Replace the substitution in the result

Replace \(u\) with \(f(x)\) in the result to obtain the final answer: $$= \frac{5}{4}f^4(x) + \frac{7}{3}f^3(x) + \frac{1}{2}f^2(x) + C$$ This is the final result of the given integral.

Key concepts

Integration by Substitution

In calculus, integration by substitution is a technique which simplifies complex integrals. It's much like the mathematical equivalent of a costume change in a play; by changing the variables, we often get a simpler expression that's easier to work with.

Imagine you have an integral that involves a function and its derivative, much like our exercise with \( f(x) \). Here, we use substitution to replace \( f(x) \) with a new variable, let's say \( u \). This change turns our complicated integral into one that is often simpler and one that we can tackle with basic integration rules.

Remember, when we perform the substitution, we need to express everything - including the differential - in terms of the new variable \( u \). So, if we let \( u = f(x) \), then the differential \( du \) corresponds to \( f'(x)dx \). After performing the integration in terms of \( u \), we'll substitute back to the original variable to get the final answer. This technique is widely used when the integrand is a product of a function and its derivative, or can be manipulated to appear that way.

Power Rule for Integration

The power rule for integration is a primary tool in the belt of calculus, simplifying the process of finding antiderivatives of power functions. It's like a basic recipe that applies to a variety of ingredients—in this case, powers of \( x \).

If you have a function \( x^n \) where \( n \) is any real number except -1, integrating it is straightforward. You simply add one to the exponent \( n \) to get \( n + 1 \) and then divide by this new exponent. Symbolically, this is expressed as: $$\int x^n dx = \frac{1}{n+1}x^{n+1} + C$$ where \( C \) is the constant of integration.

In our exercise, applying the power rule to each term separately after substituting \( u \) for \( f(x) \) allows us to find the antiderivative easily. Adjusting the power and dividing by the new exponent, we then arrive at the integrated form of each term. This technique is a quick and efficient method to integrate polynomials and other functions with variables raised to a power.

Antiderivatives

Moving onto antiderivatives, these are the opposite of derivatives. While a derivative represents the rate of change, an antiderivative gives us a function that, when differentiated, would yield the original function. It helps us in finding the original quantity when we know the rate of change.

In the context of our problem, after performing the substitution and simplifying the integral, we sought the antiderivative to find the function whose derivative would yield our integrand. This is a journey back to the origin, climbing the hill after watching how something rolls down, so to speak. Importantly, since differentiation is precise but integration can lead to many possible functions (all differing by a constant), antiderivatives are often expressed with a \( + C \) to indicate the constant of integration.

Antiderivatives are not just academic exercises; they are critical in solving real-world problems involving accumulation or reconstruction of quantities from their rates of change, like in physics for velocity and acceleration or in economics for cost and marginal cost.

Continuous Functions

The concept of continuous functions is fundamental in understanding integrals and antiderivatives. A function is continuous if you can draw it without lifting your pen from the paper—there are no breaks, jumps, or holes. Mathematically, a function \( f(x) \) is continuous at a point \( x = a \) if \( \lim_{x\to a} f(x) = f(a) \).

In the world of integrals, continuity is crucial because we can only integrate over intervals where the function is continuous. If there were gaps, we'd have to split the integral into pieces and handle each chunk separately. For our exercise, it's stated that the function \( f(x) \) and its derivatives are continuous for all real numbers, which assures us that we can proceed with our process without worrying about any mathematical potholes.

The continuity of a function is a promise of smooth sailing as we travel the curve to find the area underneath or to reverse-engineer the curve from its slope. It ensures that our integrals behave well, giving us meaningful and applicable results in pure and applied mathematics.