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Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int x^{-3}(x+1) d x$$

Short Answer

Expert verified
The antiderivative is \(-x^{-1} - \frac{1}{2}x^{-2} + C\).

Step by step solution

01

Simplify the Integrand

The given integral is \( \int x^{-3}(x+1) \ dx \). Begin by simplifying the integrand. Distribute \(x^{-3}\) to both terms in the parentheses:\[ \int (x^{-3} \cdot x + x^{-3} \cdot 1) \ dx = \int (x^{-2} + x^{-3}) \ dx \]
02

Apply Power Rule for Integration

Integrate each term separately using the power rule for integration \( \int x^n \ dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). Apply the rule to the simplified expression:- For \( x^{-2} \), the integral is \( \frac{x^{-1}}{-1} = -x^{-1} \).- For \( x^{-3} \), the integral is \( \frac{x^{-2}}{-2} = -\frac{1}{2}x^{-2} \).Combine these results:\[ \int (x^{-2} + x^{-3}) \ dx = -x^{-1} - \frac{1}{2}x^{-2} + C \]
03

Verify by Differentiation

Differentiate the result to ensure it matches the original integrand. Start with \(-x^{-1} - \frac{1}{2}x^{-2} + C\).- The derivative of \(-x^{-1}\) is \(x^{-2}\).- The derivative of \(-\frac{1}{2}x^{-2}\) is \(-x^{-3}\).- The derivative of \(C\) is \(0\).Adding these, we get: \(x^{-2} - x^{-3}\), which simplifies to \(x^{-3}(x+1)\), confirming the integration was performed correctly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integrals
Indefinite integrals are a fundamental part of calculus, representing the general form of the antiderivative of a function. The term "indefinite" is used because these integrals do not have specified bounds. Therefore, they include a constant of integration, typically denoted as "C". This constant accounts for the infinite number of antiderivatives that can result from the process. In the context of indefinite integrals, we use the symbol \int to represent integration. This symbol, along with the function and differential (such as \(dx\)), indicates the process of finding what function will differentiate to yield the original integrand. Indefinite integrals are closely related to the reverse process of differentiation. While differentiation focuses on finding the rate of change of a function, indefinite integration aims to determine the original function from its rate of change. This makes understanding both processes crucial for mastering calculus.
Antiderivatives
The concept of antiderivatives is central to understanding and solving indefinite integrals. An antiderivative of a function is another function whose derivative results in the original function. In essence, if you have a function \(f(x)\), an antiderivative \(F(x)\) will satisfy the relationship \(F'(x) = f(x)\). This integration concept allows us to reconstruct a function from its derivative or find more general forms by integrating a function over its domain. For example, if given the function \(x^{-2} + x^{-3}\), finding the antiderivative involves integrating each component separately and adding the constant C to account for arbitrary shifts in the function. Antiderivatives are fundamentally linked to the area under a curve in a graph. Since differentiation focuses on slopes, integration focusing on reconstructing the function paves the way for numerous applications in physics, engineering, and other fields involving rates of change and accumulated quantities.
Integration Techniques
Mastering various integration techniques is essential to solve more complex integrals. For our original exercise, we used the power rule for integration as our primary technique. The power rule states that to integrate a term of the form \(x^n\), where n is not equal to -1, the result is \(\frac{x^{n+1}}{n+1} + C\). This powerful rule helps in directly applying to terms where the function is polynomial-like, providing a straightforward path from integration to antiderivative.Besides the power rule, there are other techniques such as substitution, integration by parts, and partial fraction decomposition. Each technique has its place and importance depending on the form of the integrand. In practice, recognizing the opportunity to simplify or manipulate an integral is crucial for selecting the correct method and achieving a correct and efficient solution.Our exercise also demonstrated the simplification technique by disttributing and breaking down an integrand before integrating. These strategies lead to more manageable calculations, making the problem easier to solve.
Verification by Differentiation
Verifying an indefinite integral solution through differentiation is a reliable method to confirm its correctness. After obtaining an antiderivative, differentiating it should return the original integrand. This process ensures that integration was done accurately with no mistakes in calculation or adjustments.In our solution, after finding the antiderivative \(-x^{-1} - \frac{1}{2}x^{-2} + C\), taking its derivative yielded \(x^{-2} - x^{-3}\). This confirmed the solution was correct, as it matched the simplified form of the original is integrand \(x^{-3}(x+1)\).This reverse check serves as an additional layer of assurance, fostering a deep understanding of the intimate relationship between derivatives and integrals. In hallmarks of calculus learning, being able to backtrack your steps and validate your answers rounds out your problem-solving skills and confidence in handling complex equations.

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Most popular questions from this chapter

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