Evaluating $\int_3^7x^3dx$ using $\sum_{j=1}^Nj^3=\left(\frac12{N(N+1)}\right)^2$ Question: 
$$\int_3^7x^3dx$$
I know how to solve it using the Power Rule, but I wanted to know how to solve it using:
$$1^3+2^3+\cdots+N^3=\sum_{j=1}^Nj^3=\left({N(N+1)\over2}\right)^2$$
So I'm just confused on how we use that formula to find an answer to that integral. Thank you!
 A: Recall $$\int_a^bf(x)dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x$$ 
where $\Delta x=\frac{b-a}{n}$ and $x_i=a+i\Delta x$ (I'm using right-endpoints in our Riemann-sum). Thus $$\int_3^7x^3dx=\lim_{n\to\infty}\sum_{i=1}^n\left(3+\frac{4i}{n}\right)^3\cdot\frac{4}{n}$$ $$=\lim_{n\to\infty}\frac{4}{n}\sum_{i=1}^n\left(3^3+3\cdot3^2\cdot\frac{4i}{n}+3\cdot3\cdot\frac{(4i)^2}{n^2}+\frac{(4i)^3}{n^3}\right)$$ $$=\lim_{n\to\infty}\frac{4}{n}\sum_{i=1}^n\left(27+\frac{108}{n}\cdot i+\frac{144}{n^2}\cdot i^2+\frac{64}{n^3}\cdot i^3\right)$$ $$=\lim_{n\to\infty}\left(\frac{4}{n}\sum_{i=1}^n27+\frac{432}{n^2}\sum_{i=1}^n i+\frac{576}{n^3}\sum_{i=1}^ni^2+\frac{256}{n^4}\sum_{i=1}^n i^3\right)$$
Then note $$\sum_{i=1}^nc=nc,\space\space \sum_{i=1}^n i=\frac{n(n+1)}{2},\space\space \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6},\space\space\sum_{i=1}^ni^3=\left(\frac{n(n+1)}{2}\right)^2$$ Thus $$\int_3^7x^3dx=$$ $$=\lim_{n\to\infty}\left(\frac{4}{n}\cdot27n+\frac{432}{n^2}\cdot\frac{n(n+1)}{2}+\frac{576}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{256}{n^4}\cdot\left(\frac{n(n+1)}{2}\right)^2\right)$$ $$=\lim_{n\to\infty}\left(108+216\left(1+\frac{1}{n}\right)+96\left(2+\frac{3}{n}+\frac{1}{n^2}\right)+64\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\right)$$ $$=108+216+96\cdot2+64=580$$
A: Hint
You wouldn't really be able to solve the integral exactly, but you can get a decent estimation. Think about Riemann Sums.
