Problem 38 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indefinite integral. $$\int \frac{x^{4}-1}{x^{2}} d x$$

Step-by-Step Solution

Answer

The indefinite integral of the given function is: \(\int \frac{x^4 - 1}{x^2} dx = \frac{x^3}{3} + x + C\).

Step 1: Simplify the given function

Divide each term of the numerator by the denominator i.e. \(x^2\). Here we go: \[\frac{x^4 - 1}{x^2} = \frac{x^4}{x^2} - \frac{1}{x^2}\] Now simplify each term, we get \[= x^2 - x^{-2}\] The given function is simplified and we can now proceed to find its indefinite integral.

Step 2: Calculate the indefinite integral

To calculate the indefinite integral of the simplified function, we need to integrate each term separately: \[\int (x^2 - x^{-2}) dx\] Using the power rule for integration, we get \[\int x^2 dx - \int x^{-2} dx\] The power rule for integration states that \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where n is a constant. We'll apply this rule to both terms: \[\frac{x^3}{3} - \frac{x^{-1}}{-1} + C\] Now, simplify the expression: \[\frac{x^3}{3} + x + C\] We have now found the indefinite integral. The indefinite integral of the given function is: \[\int \frac{x^4 - 1}{x^2} dx = \frac{x^3}{3} + x + C\]

Key Concepts

Integration Techniques

When faced with finding the indefinite integral of a function, knowing which integration technique to apply is crucial. In our example, the original expression, \( \int \frac{x^{4}-1}{x^{2}} \, dx \), can be intimidating. However, by recognizing a simplification technique, we reduce the problem to something manageable through integration.

By simplifying the given rational expression before integrating, we effectively make the problem easier to solve. Common techniques include:

• Substitution
• Integration by parts
• Partial fraction decomposition
• Simplification, as used here

In this case, simplification allowed us to transform a complex fraction into a combination of simpler polynomial terms, enabling straightforward application of other rules, like the power rule.

Power Rule

The power rule is one of the most basic yet powerful techniques used in integration. It states that for any real number \( n eq -1 \), the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

In our solution, once we simplified the expression \( \frac{x^4 - 1}{x^2} \) into \( x^2 - x^{-2} \), we applied the power rule to each term separately:

• The integral of \( x^2 \) becomes \( \frac{x^3}{3} \)
• The integral of \( -x^{-2} \) becomes \( -\left(-\frac{1}{x}\right) \), which simplifies to \( x \)

By seamlessly using the power rule, we calculated each term's integral, resulting in the succinct expression \( \frac{x^3}{3} + x + C \). This showcases the straightforward yet essential nature of the power rule in integration.

Simplifying Rational Expressions

Simplifying rational expressions is often a necessary step before integration. It involves breaking down a complex fraction into a simpler form that can be more readily integrated. The method used in this exercise involved dividing each term of the numerator by the denominator.

Here's how we did it:

• The initial expression was \( \frac{x^4 - 1}{x^2} \)
• We separate this into two simpler fractions: \( \frac{x^4}{x^2} \) and \( -\frac{1}{x^2} \)
• Each fraction is then simplified to \( x^2 \) and \( -x^{-2} \) respectively.

This step is crucial because it transforms a potentially difficult integration problem into an easier one. Just remember, simplification is all about making life easier, allowing you to apply straightforward integration techniques like the power rule.

Calculus

Calculus is the branch of mathematics that deals with continuous change. It is divided into two main parts: differential calculus and integral calculus. Integral calculus focuses on finding functions' integrals, which can mean finding areas under curves or solving differential equations.

In this exercise, we dealt with indefinite integrals, which provide a family of functions that could be solutions to differential equations. These integrals do not have set limits, unlike definite integrals, which compute a number representing the area under a curve between two points.

• Indefinite integrals involve a constant \( C \) as part of the general solution.
• The purpose of calculating an indefinite integral is often to "reverse" the process of differentiation.

Mastering these principles expands your ability to tackle real-world problems involving rates of change and accumulated quantities, fundamental concepts in calculus.