I'm working through a question and seem to have hit a brick wall with my manipulation. I know I am right up to the point where I am stuck but I will show my working anyway.
My question
By choosing a suitable dissection of $[1, 3]$ ,show that $\int^3_1 2x^2 -1 dx$ exists and find its value.
My suitable dissection is $[1,1+\frac{2}{n},1+\frac{4}{n}...,1+\frac{2n}{n}]$
From there I started working out the upper sum $s^u$ and the lower sum $s_l$ ,
$s_l$ = $\sum_{i=1}^n (2(\frac{2i-2}{n})^2-1)\frac{2}{n}$
After manipulation I got to
$16\sum^n_{i=1} \frac{i^2-2i+1}{n^3}+\frac{i-1}{n^2}+\frac{2}{n}$
and fo the upper sum
$s^u$= $\sum_{i=1}^n (2(1+\frac{2i}{n})^2-1)\frac{2}{n}$
After manipulation I got to:
$16 \sum_{i=1}^{n}\frac{i^2}{n^3}+\frac{i}{n^2}+\frac{2}{n}$
I then have been given the tip that $\sum^n_{i=1} i^2 = \frac{n^3}{3}+ \frac{n^2}{2}+ \frac{n}{6}$
There is probably a blatant solution here but I am unable to see it, help would be greatly appreciated.