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?
$\displaystyle \begin{aligned}\lim_{n\to\infty}\sum_{k=1}^{n}\lef t\{{\color{red}a}+\frac{(b-a)k}{n}\right\}\cdot\frac{b-a}{n}&=\lim_{n\to\infty}\sum_{k=1}^{\infty}\left\{ \frac{a(b-a)}{n}+\frac{(b-a)^2k}{n^2}\right\}\\Yes, thank you.
But now I'm still stuck... How would I go on with this problem?
&=\lim_{n\to\infty}\left\{a(b-a)+\frac{(b-a)^2n(n+1)}{2n^2}\right\}\quad{\color{red}\star}\\
&=ab-a^2+\frac{b^2}{2}-ab+\frac{a^2}{2}\\
&=\frac{b^2-a^2}{2}\end{aligned}$
$\displaystyle {\color{red}\star}~\sum_{k=1}^{n}k=\frac{n(n+1)}{2 }~~\sum_{k=1}^{n}c=cn$