Find Antiderivative F(x) With F(1)=0

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of calculus, specifically focusing on a cool problem involving antiderivatives. You know, those functions that are the opposite of derivatives? They're super important for solving all sorts of problems, from finding areas under curves to understanding motion. We've got a specific function here, f(x)=rac{4}{x^2}-rac{10}{x^6}, and we need to find its antiderivative, which we'll call $F(x)$. But here's the twist: we're not just looking for any antiderivative; we need the specific one that satisfies a condition, $F(1)=0$. This condition is crucial because, as you'll see, antiderivatives have a constant of integration that can take on any value. This condition helps us nail down that constant and get our unique $F(x)$. Let's break down how we get there step-by-step, making sure we understand each part. We'll start by rewriting our function $f(x)$ in a more calculus-friendly format, using negative exponents. This makes applying the power rule for integration a breeze. Remember the power rule? It states that the integral of $x^n$ is rac{x^{n+1}}{n+1}, provided $n eq -1$. We'll apply this rule to each term in our function separately. After we integrate, we'll add our constant of integration, typically denoted by 'C'. Then, we'll use the given condition $F(1)=0$ to solve for C. This is where the specific part of the problem comes into play. It's like finding a specific key that fits a lock, rather than just any key. So, grab your thinking caps, and let's get this calculus party started! We'll aim to make this as clear and as fun as possible, because math should be, right?

Understanding Antiderivatives and the Power Rule

Alright, let's get down to business with our function f(x)=rac{4}{x^2}-rac{10}{x^6}. Before we can find the antiderivative $F(x)$, it's super helpful to rewrite $f(x)$ using negative exponents. This is because the power rule for integration is most easily applied when the variable is raised to a power. So, rac{4}{x^2} becomes $4x^{-2}$ and rac{10}{x^6} becomes $10x^{-6}$. Our function now looks like this: $f(x) = 4x^{-2} - 10x^{-6}$. Now, let's talk about the power rule for integration. If you have a term like $ax^n$, its antiderivative is rac{a}{n+1}x^{n+1}, as long as $n eq -1$. This is the magic formula we'll use. We apply this rule to each term in our function $f(x)$ individually. For the first term, $4x^{-2}$, our 'a' is 4 and our 'n' is -2. Applying the power rule, the antiderivative of $4x^{-2}$ is rac{4}{-2+1}x^{-2+1} = rac{4}{-1}x^{-1} = -4x^{-1}. For the second term, $-10x^{-6}$, our 'a' is -10 and our 'n' is -6. Applying the power rule here gives us rac{-10}{-6+1}x^{-6+1} = rac{-10}{-5}x^{-5} = 2x^{-5}. So, the general antiderivative of $f(x)$ is the sum of these results, plus our constant of integration, 'C'. This gives us $F(x) = -4x^{-1} + 2x^{-5} + C$. Remember, this 'C' represents any constant value. Without it, we'd only have one antiderivative, but in reality, there's an infinite family of antiderivatives, all differing by this constant. It's like having a whole set of parallel lines, all with the same slope but different y-intercepts.

Applying the Condition F(1)=0 to Find C

Now, here's where we make our antiderivative specific. We are given the condition that $F(1)=0$. This means that when we plug in $x=1$ into our general antiderivative $F(x) = -4x^{-1} + 2x^{-5} + C$, the result must be 0. Let's substitute $x=1$ into our expression for $F(x)$: $F(1) = -4(1)^{-1} + 2(1)^{-5} + C$. Since any power of 1 is just 1 (even negative powers!), this simplifies to: $F(1) = -4(1) + 2(1) + C = -4 + 2 + C$. We know that $F(1)$ must equal 0, so we set up the equation: $0 = -4 + 2 + C$. Simplifying the right side, we get $0 = -2 + C$. To solve for C, we simply add 2 to both sides of the equation: $C = 2$. Awesome! We've found the value of our constant of integration. This means our specific antiderivative, the one that satisfies $F(1)=0$, is $F(x) = -4x^{-1} + 2x^{-5} + 2$. It's important to check our work. Does this $F(x)$ satisfy the condition $F(1)=0$? Let's plug in $x=1$: $F(1) = -4(1)^{-1} + 2(1)^{-5} + 2 = -4(1) + 2(1) + 2 = -4 + 2 + 2 = 0$. Yes, it does! That's fantastic. Now, if we want, we can rewrite $F(x)$ back in terms of fractions, like the original function $f(x)$ was presented. So, $-4x^{-1}$ becomes rac{-4}{x} and $2x^{-5}$ becomes rac{2}{x^5}. Our final, specific antiderivative is F(x) = -rac{4}{x} + rac{2}{x^5} + 2. This is the unique function whose derivative is $f(x)$ and passes through the point $(1, 0)$ on its graph. Pretty neat, huh?

Finalizing the Antiderivative F(x)

So, we've successfully navigated the path to finding our specific antiderivative $F(x)$. We started with the function f(x)=rac{4}{x^2}-rac{10}{x^6}, rewrote it as $f(x) = 4x^{-2} - 10x^{-6}$ to make integration easier. Using the power rule for integration, $\int ax^n dx = \frac{a}{n+1}x^{n+1} + C$, we found the general antiderivative: $F(x) = -4x^{-1} + 2x^{-5} + C$. The 'C' here is our constant of integration, which represents the family of all possible antiderivatives. To pinpoint the exact antiderivative we needed, we used the condition $F(1)=0$. By substituting $x=1$ into our general $F(x)$ and setting the result to zero, we solved for C and found that $C=2$. This gave us our unique antiderivative: $F(x) = -4x^{-1} + 2x^{-5} + 2$. Finally, to present our answer in a form similar to the original function, we converted the negative exponents back into fractional form. Thus, the antiderivative of f(x)=rac{4}{x^2}-rac{10}{x^6} with the condition $F(1)=0$ is F(x) = -rac{4}{x} + rac{2}{x^5} + 2. This is a key concept in calculus, showing how an initial condition can specify a unique solution from an infinite family of possibilities. It's like giving a car a destination; without it, the car could go anywhere, but with it, it's on a specific journey. Keep practicing these, guys, because the more you do, the more natural they become! We'll see you in the next article for more math adventures!