Problem 21 - Applied Calculus

Question

Find an antiderivative. $$ f(x)=x^{6}-\frac{1}{7 x^{6}} $$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = \frac{x^7}{7} + \frac{1}{35x^5} + C \).

Step 1: Identify the function components

The given function is \( f(x) = x^6 - \frac{1}{7x^6} \). It consists of two parts: the term \( x^6 \) and the term \( -\frac{1}{7x^6} \). We will find the antiderivative of each part separately.

Step 2: Find the antiderivative of the first term

For the term \( x^6 \), the antiderivative is found by increasing the exponent by one and dividing by the new exponent. So, the antiderivative of \( x^6 \) is \( \frac{x^7}{7} \).

Step 3: Find the antiderivative of the second term

The second term is \( -\frac{1}{7x^6} = -\frac{1}{7} x^{-6} \). Applying the power rule for integration, we increase the exponent by one \( x^{-6 + 1} \), resulting in \( -5 \), and divide by the new exponent. Thus, the antiderivative is \( \frac{-1}{7(-5)}x^{-5} = \frac{1}{35x^5} \).

Step 4: Combine the antiderivatives

Combine the antiderivatives of each term. The antiderivative of the entire function is: \[ F(x) = \frac{x^7}{7} + \frac{1}{35x^5} + C \] where \( C \) is the constant of integration.

Key Concepts

Power Rule for Integration

The Power Rule for Integration is a straightforward method to find the antiderivative, or the integral, of polynomial functions. Simply put, this rule is applied to functions of the form \( x^n \), where \( n \) is any real number except \(-1\). To apply the Power Rule:

• Add 1 to the exponent: if you have \( x^n \), it becomes \( x^{n+1} \).
• Divide by the new exponent: take the answer and divide it by \( n+1 \).

After this process, don't forget to add the constant of integration \( C \) at the end, which is very important and we'll cover later. For example, in the given problem, the term \( x^6 \) was integrated to become \( \frac{x^7}{7} \) by using this rule. Similarly, the second term \(-\frac{1}{7x^6} = -\frac{1}{7}x^{-6}\) uses the Power Rule to become \( \frac{1}{35x^5} \). Here, increasing the exponent by 1 and then dividing by the new exponent can simplify terms like this, which might initially appear tricky.

Constant of Integration

The Constant of Integration emerges when computing indefinite integrals. It's represented by the symbol \( C \) in your final antiderivative expression. This constant accounts for the fact that there are infinitely many antiderivatives for any given function. Each antiderivative is a vertical shift of another, and the constant \( C \) represents this shift.For indefinite integrals, omitting the constant of integration can lead to incomplete solutions. Including \( C \), as part of the integration process, acknowledges the entirety of possible solutions. So, whenever you integrate a function, remember to add \( C \) at the end.In our exercise, after finding the antiderivative of each term in the function, we combined them to get \( F(x) = \frac{x^7}{7} + \frac{1}{35x^5} + C \). Adding \( C \) ensures that we're accounting for all possible antiderivative solutions.

Integration of Polynomial Functions

Polynomial functions are among the simplest to integrate due to their straightforward rules. These functions are expressions that consist of variables raised to whole number powers, such as \( x^2 \), \( x^3 \), or a sum of such terms. To integrate polynomial functions, each term is treated separately.Let's break down the steps:

• First, apply the Power Rule to each polynomial term, as covered in the previous sections. Increase each exponent by 1 and divide by the new exponent.
• After integrating each term individually, combine them. This involves adding or subtracting terms to form the integrated polynomial.
• Finally, remember to include the constant of integration \( C \) to account for the family of possible solutions.

The exercise demonstrates this by integrating the polynomial function \( f(x) = x^6 - \frac{1}{7x^6} \). Each term was integrated separately, and then combined to form the final antiderivative. This technique makes dealing with more complex polynomials a breeze, as breaking them down into simpler parts simplifies the integration process.