Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the general antiderivative \(F(x)+C\) for each of the following. $$ f(x)=\frac{x^{6}-x}{x^{3}} $$

Step-by-Step Solution

Answer

The general antiderivative is \( F(x) = \frac{x^4}{4} + \frac{1}{x} + C \).

Step 1: Simplify the Function

First, simplify the given function by dividing each term of the numerator by the denominator: \[ f(x) = \frac{x^6}{x^3} - \frac{x}{x^3} = x^3 - x^{-2} \]

Step 2: Integrate Each Term

Now, integrate each term separately to find the antiderivative. Use the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \):\[ \begin{align*} \int (x^3 - x^{-2}) \, dx &= \int x^3 \, dx - \int x^{-2} \, dx \ &= \frac{x^{4}}{4} - \frac{x^{-1}}{-1} = \frac{x^{4}}{4} + x^{-1} \end{align*} \]

Step 3: Combine the Results

Combine the results from the previous step to express the general antiderivative as a single expression:\[ F(x) = \frac{x^4}{4} + \frac{1}{x} + C \]

Step 4: Add the Constant of Integration

Add the constant \( C \) to account for the constant of integration in the antiderivative:\[ F(x) = \frac{x^4}{4} + \frac{1}{x} + C \]

Key Concepts

Power Rule for Integration

The Power Rule for Integration is a fundamental method used in calculus to find antiderivatives, or indefinite integrals. This rule allows us to integrate terms of the form \(x^n\), where \(n\) is any real number except \(-1\). The rule states that to integrate \(x^n\), you increase the exponent by one, then divide by the new exponent. Mathematically, it is expressed as: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] This approach simplifies the process of finding antiderivatives because the operation is systematic. For example, to find the antiderivative of \(x^3\) using the power rule, you increase the exponent to 4 and divide by 4, resulting in \(\frac{x^4}{4}\).

• The power rule is applicable as long as \(n\) is not \(-1\), which would create a division by zero.
• For \(n = -1\), the antiderivative is a logarithmic function, specifically \(\ln|x|\).

In the exercise, you are using the power rule to integrate each term like \(x^3\) and \(x^{-2}\), applying the rule separately to each term.

Calculus Problem Solving

Solving calculus problems, such as finding antiderivatives, involves a series of logical steps and the application of specific rules and techniques. In the given exercise, you begin by simplifying the expression, which makes it easier to integrate. The original function, \(f(x)=\frac{x^6-x}{x^3}\), can look complex at first glance. In problem-solving:

• First, simplify the expression wherever possible. Divide or factor out terms to make the equation manageable. In this case, dividing each term in the numerator by the denominator.
• Next, apply integration techniques, such as the power rule, to each simplified term separately.
• Finally, combine the integrated terms and include the constant of integration, \(C\), which accounts for unknown constants in indefinite integrals.

These steps ensure that each part of the problem is systematically addressed, minimizing mistakes and clarifying the process of finding the antiderivative.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a key skill in calculus that makes complex functions easier to work with. When you simplify expressions, you're breaking them down into forms that are straightforward to apply calculus rules to. For example, the function \(f(x)=\frac{x^6-x}{x^3}\) initially seems complicated due to the division. To simplify, divide each term in the numerator by the denominator:

• Divide \(x^6\) by \(x^3\) to get \(x^3\). This reduces power and complexity.
• Divide \(x\) by \(x^3\), resulting in \(x^{-2}\). This term becomes manageable and ready for integration.

By simplifying the expression from \(\frac{x^6-x}{x^3}\) to \(x^3 - x^{-2}\), you prepare the function for straightforward integration. This simplification allows you to apply the power rule more effectively, turning the task of finding an antiderivative into a series of easier steps.