I can't quite get the upper and lower sums of $f(x)=x^2$ from $[a,b]$ (where we are defining the function) .
Here I'm talking about Darboux sums of the form $U(f, P_n) = \displaystyle\sum_{i=1}^n M_i(t_i-t_{i-1})$ and $L(f,P_n)=\displaystyle\sum_{i=1}^n m_i(t_i-t_{i-1}) $ where $m_i=inf\{f(x): t_{i-1} \le x \le t_i\}$ and $M_i=sup\{f(x): t_{i-1} \le x \le t_i\}$. (Also, we assume a partition $P_n=\{t_{\mathbb{0}},t_1,t_2,...t_n\}$)
I was thinking that $t_i= \frac{i(b-a)}{n}$, but if we do this in the end the sum is multiplied by $\frac{(b-a)^3}{n^3}$ which is a bit different from the expected $\frac{b^3-a^3}{n^3}$ that easy way to calculate the integral yields. (where the $\frac {1}{n^3}$ cancels out with the limit of the sum that ends up as $\displaystyle\sum_{i=1}^n=i^2$).
So if anyone could shine some light on how to work this out (pick a better $t_i$ or if I'm doing the sum incorrectly) it would be very much appreciated thanks.