Let $P$ be a partition of $[0, b]$ defined as $P = \{ 0 = x_0 < x_1 < > \ldots < x_n = b\}$, and let $c_i \in [x_{i-1}, x_i]$ for every $1 > \leq i \leq n$.
Prove: $$\left|\Sigma^n_{i=1}c_i\Delta x_i - \frac{b^2}{2}\right|\leq \frac{1}{2}\Sigma^n_{i=1}(\Delta x_i)^2$$
Here's my attempting, trying to go from left to right:
Using the triangle inequality: $$\left|\Sigma^n_{i=1}c_i\Delta x_i +\left(- \frac{b^2}{2}\right)\right| \leq \left|\Sigma^n_{i=1}c_i\Delta x_i\right| + \left|-\frac{b}{2}\right|\\ = \Sigma^n_{i=1}c_i\Delta x_i + \frac{b}{2}=\Sigma^n_{i=1}c_i\Delta x_i + \frac{\left(\Sigma^n_{i=1}\Delta x_i\right)^2}{2} \\\leq \Sigma^n_{i=1}x_i \cdot \Delta x_i + \frac{\left(\Sigma^n_{i=1}\Delta x_i\right)^2}{2}$$
This is as far as I could develop it. Any way to proceed?
Hint \begin{equation} \frac{1}{2}\sum_1^n (\Delta x_i)^2 = -\frac{b^2}{2} + \sum_1^n x_i \Delta x_i \end{equation} and \begin{equation} \sum_1^n x_i \Delta x_i = \sum_1^n (\Delta x_i)^2 + \sum_1^n x_{i-1} \Delta x_i \end{equation}