Finding The Antiderivative Of F(x) = 10/x^2 - 10/x^5 With F(1) = 0
Introduction
In the realm of calculus, finding antiderivatives is a fundamental operation with wide-ranging applications. This article delves into the process of determining the antiderivative of the function f(x) = 10/x² - 10/x⁵, a quintessential problem that highlights the core principles of integration. We'll not only compute the antiderivative but also explore the significance of the given initial condition, F(1) = 0, in uniquely defining the antiderivative function. This constant of integration, often denoted as 'C', plays a crucial role in tailoring the general antiderivative to a specific solution that satisfies the initial value problem. Our journey will take us through the power rule of integration, the manipulation of exponents, and the careful application of the initial condition to arrive at the definitive form of F(x). This exploration serves as a stepping stone for understanding more complex integration techniques and their applications in various fields, including physics, engineering, and economics. We'll break down the problem step-by-step, ensuring clarity and a comprehensive understanding of the underlying concepts. This exercise is not just about finding the answer; it's about mastering the process of integration and appreciating its power in solving real-world problems. Understanding antiderivatives is crucial for grasping concepts like displacement from velocity, total revenue from marginal revenue, and many other applications where the rate of change is known, and the original function needs to be determined. So, let's embark on this mathematical journey and unravel the antiderivative of the given function, keeping in mind the elegance and precision that calculus offers.
Understanding the Function f(x) = 10/x² - 10/x⁵
Before we embark on the integration process, it's crucial to thoroughly understand the function we're dealing with: f(x) = 10/x² - 10/x⁵. This function is a combination of two terms, each involving a power of x in the denominator. Recognizing this structure is key to applying the power rule of integration effectively. The function can be rewritten using negative exponents as f(x) = 10x⁻² - 10x⁻⁵. This transformation is not merely cosmetic; it's a strategic step that allows us to directly apply the power rule, which states that the integral of xⁿ is (xⁿ⁺¹)/(n+1), provided n ≠ -1. Understanding the nuances of exponents and their manipulation is fundamental to calculus, and this function provides an excellent opportunity to reinforce that understanding. The coefficients '10' in each term are constants that can be factored out of the integral, simplifying the process. This is another crucial property of integrals: the integral of a constant times a function is the constant times the integral of the function. By breaking down the function into its constituent terms and understanding the properties of exponents and integrals, we set the stage for a smooth and accurate integration process. This preliminary analysis is not just a formality; it's a critical step in problem-solving that often leads to a deeper understanding of the underlying mathematical structure. Furthermore, recognizing the domain of the function is essential. Since we have x in the denominator, x cannot be zero. This observation will become relevant when we consider the constant of integration and the initial condition. A solid grasp of the function's behavior and properties lays the groundwork for a successful application of integration techniques. So, with a clear understanding of f(x) = 10/x² - 10/x⁵, we are well-prepared to find its antiderivative.
Finding the Antiderivative F(x)
Now, let's dive into the core of the problem: finding the antiderivative, F(x), of f(x) = 10x⁻² - 10x⁻⁵. As we established earlier, the power rule of integration is our primary tool here. The power rule dictates that ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration. Applying this rule to each term in f(x), we get: ∫10x⁻² dx = 10∫x⁻² dx = 10(x⁻¹)/(-1) + C₁ = -10x⁻¹ + C₁ and ∫-10x⁻⁵ dx = -10∫x⁻⁵ dx = -10(x⁻⁴)/(-4) + C₂ = (10/4)x⁻⁴ + C₂ = (5/2)x⁻⁴ + C₂. Combining these results, we obtain the general antiderivative: F(x) = -10x⁻¹ + (5/2)x⁻⁴ + C, where C = C₁ + C₂ is the combined constant of integration. This 'C' is a crucial element, representing the family of functions that have the same derivative, f(x). The constant of integration highlights a fundamental aspect of antiderivatives: there are infinitely many functions that differ only by a constant term and have the same derivative. To uniquely determine F(x), we need additional information, which is precisely what the initial condition F(1) = 0 provides. Before we apply the initial condition, it's beneficial to rewrite F(x) in a more conventional form without negative exponents: F(x) = -10/x + 5/(2x⁴) + C. This form makes it easier to visualize the behavior of the function and to perform substitutions. With the general antiderivative in hand, we are now ready to utilize the given condition to pinpoint the specific antiderivative that satisfies the problem's requirements. The process of finding the antiderivative involves not just applying the power rule but also understanding the significance of the constant of integration and the need for additional information to determine its value. So, with F(x) = -10/x + 5/(2x⁴) + C in our grasp, we proceed to the final step of solving for 'C'.
Applying the Initial Condition F(1) = 0
To pinpoint the specific antiderivative, we now leverage the initial condition F(1) = 0. This condition states that when x = 1, the value of the antiderivative F(x) is 0. Substituting these values into our general antiderivative, F(x) = -10/x + 5/(2x⁴) + C, we get: 0 = -10/1 + 5/(2(1)⁴) + C. Simplifying this equation, we have 0 = -10 + 5/2 + C. To solve for C, we can rearrange the equation: C = 10 - 5/2. Finding a common denominator, we get C = 20/2 - 5/2 = 15/2. Therefore, the constant of integration, C, is equal to 15/2. This value is what distinguishes our specific antiderivative from the infinite family of antiderivatives that differ only by a constant. Now that we have determined C, we can substitute it back into the general antiderivative to obtain the unique solution that satisfies the given initial condition. The initial condition acts as a filter, selecting the one antiderivative that passes through the specified point (1, 0). This process highlights the power of initial conditions in solving differential equations and related problems. By applying the condition F(1) = 0, we have effectively anchored the antiderivative at a specific point, thereby removing the ambiguity introduced by the constant of integration. This step is not just a matter of plugging in values; it's a crucial aspect of problem-solving that connects the general solution to a particular context. With C = 15/2 in hand, we are just one step away from presenting the final, complete antiderivative function. So, let's proceed to substitute this value and present the final solution.
The Final Solution: F(x) = -10/x + 5/(2x⁴) + 15/2
Having determined the constant of integration, C = 15/2, we can now present the final, specific antiderivative of f(x) = 10/x² - 10/x⁵ that satisfies the condition F(1) = 0. Substituting C into our general antiderivative, F(x) = -10/x + 5/(2x⁴) + C, we obtain: F(x) = -10/x + 5/(2x⁴) + 15/2. This is the definitive form of the antiderivative function, uniquely defined by the given initial condition. This final solution encapsulates the entire process, from recognizing the structure of the original function to applying the power rule of integration and, finally, using the initial condition to resolve the constant of integration. The function F(x) = -10/x + 5/(2x⁴) + 15/2 represents the accumulation of the rate of change described by f(x), starting from the point where F(1) = 0. This understanding is crucial in applications where antiderivatives are used to model real-world phenomena. For instance, if f(x) represents the velocity of an object, then F(x) represents its position as a function of time, with the initial condition providing the object's position at a specific time. The final solution is not just a mathematical expression; it's a powerful tool for making predictions and understanding the behavior of systems. Presenting the solution in a clear and concise manner is also important for effective communication. The form F(x) = -10/x + 5/(2x⁴) + 15/2 is readily understandable and allows for easy evaluation at different values of x. So, with confidence, we declare that the antiderivative of f(x) = 10/x² - 10/x⁵ with F(1) = 0 is indeed F(x) = -10/x + 5/(2x⁴) + 15/2. This completes our exploration of this integration problem.
Conclusion
In conclusion, we have successfully navigated the process of finding the antiderivative of the function f(x) = 10/x² - 10/x⁵ subject to the initial condition F(1) = 0. This journey has reinforced several key concepts in calculus, including the power rule of integration, the significance of the constant of integration, and the crucial role of initial conditions in determining a unique solution. We began by understanding the structure of f(x), recognizing it as a combination of power functions that could be readily integrated using the power rule. We then applied the power rule to find the general antiderivative, F(x) = -10/x + 5/(2x⁴) + C, where C represents the constant of integration. The constant of integration highlights the family of functions that share the same derivative, emphasizing that the antiderivative is not a single function but a set of functions differing only by a constant term. To pinpoint the specific antiderivative, we utilized the given initial condition, F(1) = 0. By substituting x = 1 and F(1) = 0 into the general antiderivative, we solved for C, obtaining C = 15/2. This value was then substituted back into the general antiderivative to yield the final solution: F(x) = -10/x + 5/(2x⁴) + 15/2. This final solution is not just an answer; it's a testament to the power of calculus in solving problems that arise in various fields. The process of finding antiderivatives is fundamental to understanding accumulation, rates of change, and many other essential concepts. This exercise has not only provided a specific solution but also strengthened our understanding of the broader principles of integration. As we move forward in our exploration of calculus, the skills and insights gained from this problem will undoubtedly serve as a solid foundation for tackling more complex challenges. The ability to find antiderivatives is a cornerstone of calculus, and this example has provided a valuable opportunity to hone that skill.
