Unveiling The Antiderivative: A Step-by-Step Guide

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Hey everyone! Today, we're diving into the fascinating world of calculus to find the antiderivative of a function. Specifically, we're tackling the antiderivative of 5x2βˆ’5x4x5{ \frac{5x^2 - 5x^4}{x^5} } with respect to x{ x }, keeping in mind that x{ x } can't be zero. Don't worry if it sounds a bit intimidating; we'll break it down into easy-to-follow steps. This is a fundamental concept, and understanding it will be super helpful as you explore more advanced topics in mathematics. So, grab your pens and let's get started. Finding antiderivatives is a crucial skill in calculus. Let's see how we can nail this one together!

Simplifying the Function: The First Step Towards Antidifferentiation

Alright guys, the first thing we want to do is make our lives a little easier by simplifying the given function. Currently, we have 5x2βˆ’5x4x5{ \frac{5x^2 - 5x^4}{x^5} }. It looks a bit messy, right? Let's use some algebra to clean it up. We can split the fraction into two separate terms. This gives us:

5x2x5βˆ’5x4x5{ \frac{5x^2}{x^5} - \frac{5x^4}{x^5} }

Now, we can simplify each term individually using the properties of exponents. Remember that when dividing powers with the same base, you subtract the exponents. For the first term, 5x2x5{ \frac{5x^2}{x^5} }, we subtract the exponent of x{ x } in the denominator from the exponent of x{ x } in the numerator: 2βˆ’5=βˆ’3{ 2 - 5 = -3 }. This simplifies to 5xβˆ’3{ 5x^{-3} }. Similarly, for the second term, 5x4x5{ \frac{5x^4}{x^5} }, we have 4βˆ’5=βˆ’1{ 4 - 5 = -1 }, which simplifies to 5xβˆ’1{ 5x^{-1} }. So, our simplified function becomes:

5xβˆ’3βˆ’5xβˆ’1{ 5x^{-3} - 5x^{-1} }

Much better, right? This simplified form is way easier to work with when we start finding the antiderivative. This simplification is a key step, as it transforms a complex expression into something we can easily integrate. Simplifying the function before integration not only makes the process cleaner but also minimizes the chances of errors. It's like preparing the ground before planting seeds; you want to make sure it's fertile and ready!

This simplification step is crucial, and it's something you'll encounter frequently in calculus. It's not just about getting the right answer; it's about making the process efficient and understandable. By breaking down the problem into smaller, manageable parts, we make the entire process less daunting. Always look for opportunities to simplify your functions before diving into integration. It can save you time and potential headaches down the line. Remember, practicing these steps is what makes them stick, so don't hesitate to work through various examples.

Finding the Antiderivative: The Core of the Problem

Now that we've simplified our function to 5xβˆ’3βˆ’5xβˆ’1{ 5x^{-3} - 5x^{-1} }, we're ready to find its antiderivative. Remember, the antiderivative (also known as the indefinite integral) of a function is a function whose derivative is equal to the original function. We find the antiderivative by applying the power rule of integration. For the first term, 5xβˆ’3{ 5x^{-3} }, we increase the exponent by 1 and divide by the new exponent:

∫5xβˆ’3dx=5xβˆ’2βˆ’2=βˆ’52xβˆ’2{ \int 5x^{-3} dx = \frac{5x^{-2}}{-2} = -\frac{5}{2}x^{-2} }

For the second term, 5xβˆ’1{ 5x^{-1} }, we recall that the antiderivative of xβˆ’1{ x^{-1} } or 1x{ \frac{1}{x} } is the natural logarithm of the absolute value of x{ x }. So, we have:

∫5xβˆ’1dx=5∫1xdx=5ln⁑∣x∣{ \int 5x^{-1} dx = 5 \int \frac{1}{x} dx = 5 \ln |x| }

Combining these results, the antiderivative of 5xβˆ’3βˆ’5xβˆ’1{ 5x^{-3} - 5x^{-1} } is:

βˆ’52xβˆ’2βˆ’5ln⁑∣x∣{ -\frac{5}{2}x^{-2} - 5 \ln |x| }

However, we must not forget the constant of integration, C{ C }. This is because the derivative of a constant is always zero. Therefore, any constant added to the antiderivative will still have the same derivative. Our final answer including the +C{ + C } is:

βˆ’52xβˆ’2βˆ’5ln⁑∣x∣+C{ -\frac{5}{2}x^{-2} - 5 \ln |x| + C }

This step involves applying integration rules, specifically the power rule and the integral of 1x{ \frac{1}{x} }. Understanding and correctly applying these rules is absolutely essential. The inclusion of the constant of integration, C{ C }, is non-negotiable! Failing to include C{ C } means you're only finding a particular antiderivative, not the general one. The constant of integration, C{ C }, represents an infinite family of antiderivatives, each differing by a constant value. Each and every one of them has the same derivative!

Verification and Conclusion: Checking Your Work

To ensure our solution is correct, we can verify it by taking the derivative of our antiderivative, and it should equal the original function. Let's take the derivative of βˆ’52xβˆ’2βˆ’5ln⁑∣x∣+C{ -\frac{5}{2}x^{-2} - 5 \ln |x| + C }.

The derivative of βˆ’52xβˆ’2{ -\frac{5}{2}x^{-2} } is 5xβˆ’3{ 5x^{-3} }. The derivative of βˆ’5ln⁑∣x∣{ -5 \ln |x| } is βˆ’5xβˆ’1{ -5x^{-1} }. The derivative of C{ C } is 0.

So, the derivative of βˆ’52xβˆ’2βˆ’5ln⁑∣x∣+C{ -\frac{5}{2}x^{-2} - 5 \ln |x| + C } is 5xβˆ’3βˆ’5xβˆ’1{ 5x^{-3} - 5x^{-1} }, which is the simplified form of our original function 5x2βˆ’5x4x5{ \frac{5x^2 - 5x^4}{x^5} }. Awesome, right? This confirms that our antiderivative is correct!

In conclusion, the antiderivative of 5x2βˆ’5x4x5{ \frac{5x^2 - 5x^4}{x^5} } is βˆ’52xβˆ’2βˆ’5ln⁑∣x∣+C{ -\frac{5}{2}x^{-2} - 5 \ln |x| + C }. We started by simplifying the function, then applied the rules of integration, and finally, we verified our answer by taking its derivative. Mastering the process of finding antiderivatives is a crucial foundation for more advanced topics in calculus. Remember, practice is key. Work through several examples, and soon, you'll be finding antiderivatives like a pro! Keep up the great work, and don't hesitate to revisit these steps anytime you need a refresher. The beauty of math is in the journey of understanding and discovery. Good job, everyone!

This entire process encapsulates a fundamental concept in calculus. It’s not just about the final answer, but about the steps taken to arrive there. Understanding the power rule and the integral of 1x{ \frac{1}{x} } is crucial. Don’t forget the +C{ + C }; it’s the hallmark of indefinite integration. Keep practicing, and you’ll find that these steps become second nature. Each problem you solve is a step forward, solidifying your understanding and building your confidence. Congratulations, you've conquered another calculus problem! Keep up the awesome work!