I am trying to calculate this integral using Riemann Sums, but the answer doesn't seem right in the end. So it goes with $\int _0^ax\left(1-\frac{x}{a}\right)dx\:$
First I calculated $dx=\frac{a-0}{n}=\frac{a}{n}$ and $x_k=0+\left(dx\right)k=\frac{a}{n}\cdot k=\frac{ak}{n}$ and put in the formel $\int _0^ax\left(1-\frac{x}{a}\right)dx=\lim _{n\to \infty }\left(\sum _{k=1}^nf\left(x_k\right)dx\:\right)\:$
$=\lim _{n\to \infty }(\sum _{k=1}^n\:\left[\left(\frac{ak}{n}\right)\left(1-\frac{ak}{an}\right)\right]\cdot \frac{a}{n})$
$=\lim _{n\to \infty }(\frac{a}{n}\cdot \sum _{k=1}^n\left[\left(\frac{ak}{n}\right)\left(1-\frac{k}{n}\right)\right])$
$=\lim _{n\to \infty }\left(\frac{a}{n}\left(\sum _{k=1}^n\:\frac{ak}{n}\cdot \sum _{k=1}^n\:\left(1-\frac{k}{n}\right)\right)\right)$
$=\lim _{n\to \infty }\left(\frac{a}{n}\left[\frac{a}{n}\sum _{k=1}^nk\:\left(\sum _{k=1}^n1\:-\frac{1}{n}\sum _{k=1}^nk\:\right)\right]\right)$ (since $\sum _{i=1}^n\:k=kn$ and $\sum _{i=1}^n\:i=\frac{n\left(n+1\right)}{2}$)
$=\lim _{n\to \infty }\left(\frac{a}{n}\left(\frac{a}{n}\right)\left(\frac{n\left(n+1\right)}{2}\right)\left(n-\frac{1}{n}\cdot \frac{n\left(n+1\right)}{2}\right)\right)$
Now for the final step I get:
$\lim _{n\to \infty }\left(\frac{a^2\left(n-1\right)\left(n+1\right)}{4n}\right)$ which is $=\infty$ but $\int _0^ax\left(1-\frac{x}{a}\right)dx\:=\frac{a^2}{6}$. So what am I doing wrong here?
In general, $$\frac{a}{n}\sum_{k=1}^n\big[\frac{ak}{n}\big(1-\frac{k}{n}\big)\big]\neq\frac{a}{n}\sum_{k=1}^n\frac{ak}{n}\cdot\sum_{k=1}^n\big(1-\frac{k}{n}\big).$$ Instead, you can write $$\frac{a}{n}\sum_{k=1}^n\frac{ak}{n}\big(1-\frac{k}{n}\big)=\frac{a^2}{n^2}\sum_{k=1}^n\frac{k(n-k)}{n}=\frac{a^2}{n^2}\bigg[\sum_{k=1}^nk-\sum_{k=1}^n\frac{k²}{n}\bigg]$$ and use that $$\sum_{k=1}^n k^2 = \frac{1}{6} n (n + 1) (2 n + 1)$$ as well as $$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$