1. ## Riemann sum

Use the Archimedes-Riemann Theorem to show that for $\displaystyle 0 \leq a < b$, we have $\displaystyle \int ^b_a xdx = \frac {b^2-a^2}{2}$

I'm trying to find my sum but I can't seem to remember how to do it, any hints ,please?

2. $\displaystyle \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left\{ {\frac{{\left( {b - a} \right)k}} {n}} \right\}} \cdot \frac{{b - a}} {n} = \int_a^b x dx$

3. Yes, thank you.

But now I'm still stuck... How would I go on with this problem?

4. Originally Posted by Nacho
$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left\{ {\frac{{\left( {b - a} \right)k}} {n}} \right\}} \cdot \frac{{b - a}} {n} = \int_a^b x dx$

\displaystyle \begin{aligned}\lim_{n\to\infty}\sum_{k=1}^{n}\lef t\{{\color{red}a}+\frac{(b-a)k}{n}\right\}\cdot\frac{b-a}{n}&=\lim_{n\to\infty}\sum_{k=1}^{\infty}\left\{ \frac{a(b-a)}{n}+\frac{(b-a)^2k}{n^2}\right\}\\ &=\lim_{n\to\infty}\left\{a(b-a)+\frac{(b-a)^2n(n+1)}{2n^2}\right\}\quad{\color{red}\star}\\ &=ab-a^2+\frac{b^2}{2}-ab+\frac{a^2}{2}\\ &=\frac{b^2-a^2}{2}\end{aligned}
$\displaystyle {\color{red}\star}~\sum_{k=1}^{n}k=\frac{n(n+1)}{2 }~~\sum_{k=1}^{n}c=cn$