Integral Power Rule Step-by-step Real quick, two things: I'm sorry if my notation or terminology is incorrect, and I know what I'm asking isn't strictly necessary for my studies, but writing something out step by step helps me to understand math intuitively.
I have spent a while thinking about this and cannot figure out where to go next.
I am taking Calc I and I understand differentiation intuitively and can derive functions using the various rules provided, while also understanding why they work and being able to do the steps manually.
So, for the power rule, I can show you the why of:
$$f(x) = x^n$$
$$f'(x) = nx^{n-1}$$
I also understand that to take the antiderivative, you reverse the process. I also know why the antiderivative provides a sum. Since I can't write it out step by step, though, I am not satisfied.
How can I show, algebraically, that
$$F(x^n) = \frac{x^{n+1}}{n+1}$$
So far (and I don't know how right I am, but I think I am close), I've gotten this far on my own:
$$f(x) = x^n$$
$$\lim_{n \rightarrow \infty}\sum_{i=0}^n \left[\frac{x}{n} \cdot f\left(i\cdot\frac{x}{n}\right)\right] = F(x)$$
How do I evaluate this limit? How do I cross this threshold to reach the final form of the antiderivative? I have looked through google and this site on my own for a solution, but I can't seem to come across how to proceed.
Thanks for your time.
 A: Here is what the problem boils down to: You need to have a closed formula for summing $$\sum_{i=1}^ni^k$$ I know of such a formula for $k=1,2,3$ and a buddy of mind derived one for $k=4$, but beyond that I don't know that there is a general formula for any $k$. For $k=1$ we know $$\sum_{i=1}^ni^1 = \frac{n(n+1)}{2}$$ So if you had been interested in finding the antiderivative of $f(x)=x$, you evaluate the sum as you correctly outlined: $$\lim_{n \rightarrow \infty}\left[\sum_{i=0}^n \frac{x}{n} \cdot f\left(i\cdot\frac{x}{n}\right)\right] = \lim_{n \rightarrow \infty}\left[\sum_{i=0}^n \frac{x}{n} \cdot \frac{xi}{n}\right] \\ = \lim_{n \rightarrow \infty}\left[\frac{x^2}{n^2}\sum_{i=0}^n i\right]$$ It is valid to move the $\frac{x^2}{n^2}$ outside the sum since we are summing over $i$. Now use the closed formula above to get  $$\lim_{n \rightarrow \infty}\left[\frac{x^2}{n^2}\sum_{i=0}^n i\right] = \lim_{n \rightarrow \infty}\left[\frac{x^2}{n^2}\frac{n(n+1)}{2}\right] \\ = \frac{x^2}{2}\lim_{n \rightarrow \infty}\left[\frac{n(n+1)}{n^2} \right] \\ = \frac{x^2}{2}$$ which is clearly the desired result. Again, you can find closed form sums for $k=2,3$ and maybe $4$ somewhere out there, but unless you know a general formula for an arbitrary positive integer $k$, your integral power rule options are limited.
