O sumă Riemann

Enunț. Determinaţi suma Riemann ataşată funcţiei f:\left[a,b\right] \to \mathbb{R}, f\left(x\right)=x^2, diviziunii echidistante cu n+1 noduri a intervalului \left[a,b\right] şi sistemului de puncte intermediare \xi = \left\{ \xi_i = x_i, i=\overline{1,n} \right\}, n\in \mathbb{N}^\ast.

Cum diviziunea echidistantă are cele n+1 noduri ale intervalului \left[a,b\right] date prin

    \[x_i = a+ \frac{b-a}{n}\cdot i, \enskip i=\overline{0,n},\]

se obţine

    \[\xi_i = a+ \frac{b-a}{n}\cdot i, \enskip i=\overline{1,n},\]

şi

    \[ x_i - x_{i-1} = \frac{b-a}{n}, i=\overline{1,n}. \]

Atunci, suma Rieman este

    \begin{align*} \sigma\left(f,\Delta,\xi\right) & = \sum\limits_{i=1}^{n} f\left(\xi_i\right) \cdot \left(x_i - x_{i-1}\right)\\ &= \sum\limits_{i=1}^{n} f\left(a+ \frac{b-a}{n}\cdot i\right) \cdot \frac{b-a}{n} \\ &= \frac{b-a}{n} \cdot \sum\limits_{i=1}^{n}\left(a+ \frac{b-a}{n}\cdot i\right)^2 \\ &= \frac{b-a}{n} \cdot \sum\limits_{i=1}^{n} \left(a^2 + 2a\cdot \frac{b-a}{n} \cdot i + \left(\frac{b-a}{n}\right)^2 \cdot i^2 \right) \\ &= \frac{b-a}{n} \cdot \left( a^2 \cdot n + \frac{2a\left(b-a\right)}{n} \cdot \frac{n\left(n+1\right)}{2} + \frac{\left(b-a\right)^2}{n^2} \cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\\\sigma\left(f,\Delta,\xi\right) &= a^2\left(b-a\right) + a\left(b-a\right)^2 \cdot \frac{n+1}{n} + \left(b-a\right)^3 \cdot \frac{\left(n+1\right)\left(2n+1\right)}{6n^2}, \, n\in \mathbb{N}^\ast \end{align*}